Advanced Factoring – Math Hints

Revisiting Factoring Quadratics	Factoring and Solving with Exponents
Factoring Sum and Difference of Cubes	More Practice
Factoring and Solving with Polynomials

Factoring is extremely important in math; we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section.

Revisiting Factoring Quadratics

Earlier, we learned how to factor, or “unFOIL” a trinomial into two binomials. (Remember that FOIL stands for First-Outer-Inner-Last when multiplying two binomials together):

$ {{x}^{2}}-11x-60=\left( {x-15} \right)\left( {x+4} \right)$

Find two numbers that multiplied together are $ -60$ and added together are $ -11$.

Sometimes we have to factor out common factors, or GCFs. We always want to do this first:

$ \begin{align}6{{x}^{3}}-21{{x}^{2}}-45x&=3x\left( {2{{x}^{2}}-7x-15} \right)\\&=3x\left( {2x+3} \right)\left( {x-5} \right)\end{align}$

First, take out the greatest common factor (GCF) out, which is $ 3x$. Then, factor, or “unFOIL”, but don’t forget to bring the $ 3x$ down as part of the answer. (Remember when factoring that the inner products and outer products have to add up to the middle term, $ -7x:\,\,3x+-10x=-7x$).

And don’t forget “grouping” when we have four terms (but it doesn’t always work – we found other ways to solve in Graphing and Finding Roots of Polynomial Functions later). Again, we are working with GCF‘s to do this:

$ \displaystyle \begin{align}24xy+32{{x}^{2}}y+6z+8xz&=8xy\left( {3+4x} \right)+2z\left( {3+4x} \right)\\&=\left( {8xy+2z} \right)\left( {3+4x} \right)\end{align}$

Separate the four terms into two sets of two terms. Take out the GCF out of each of the sets of terms. Then, because of the (inverse of the) distributive property, take out the expression that’s repeated, use that as one factor, and use the coefficients of this expression as the other factor.

You can factor a difference of squares, but not a sum of squares: $ 9{{x}^{2}}-25=\left( {3x-5} \right)\left( {3x+5} \right)$. However, $ 9{{x}^{2}}+25$ is prime, or irreducible; you can’t factor with reals, but would be $ \left( {3x+5i} \right)\left( {3x-5i} \right)$ using imaginary numbers)

Factoring with Sum and Difference of Cubes

For cubic binomials (sum or difference of cubes), we can factor:

$ \begin{array}{l}{{a}^{3}}+{{b}^{3}}=\left( {a+b} \right)\left( {{{a}^{2}}-ab+{{b}^{2}}} \right)\\{{a}^{3}}-{{b}^{3}}=\left( {a-b} \right)\left( {{{a}^{2}}+ab+{{b}^{2}}} \right)\end{array}$

Signs: Same, Opposite, Always Positive

Remember: S O A P when it comes to signs of cubic factoring.

Also remember that the middle term in the trinomial factor is $ \pm \,ab$, not $ \pm \,2ab$, like in perfect square trinomials.

Here are some examples of factoring sums and differences of cubes:

Sum of Cubes	Difference of Cubes
$ 27{{y}^{3}}+1$ $ \begin{align}(a&=\sqrt[3]{{27{{y}^{3}}}}=3y,\,\,\,\,\,\,b=\sqrt[3]{{{{1}^{3}}}}=1)\\&=\left( {3y+1} \right)\left[ {{{{\left( {3y} \right)}}^{2}}-\left( {3y} \right)\left( 1 \right)+{{1}^{2}}} \right]\\&=\left( {3y+1} \right)\left( {9{{y}^{2}}-3y+1} \right)\end{align}$	$ 8{{x}^{3}}{{y}^{6}}-125{{z}^{3}}$ $ \begin{align}\,(a&=\sqrt[3]{{8{{x}^{3}}{{y}^{6}}}}=2x{{y}^{2}},\,\,\,\,\,\,b=\sqrt[3]{{125{{z}^{3}}}}=5z)\\&=\left( {2x{{y}^{2}}-5z} \right)\left[ {{{{\left( {2x{{y}^{2}}} \right)}}^{2}}+\left( {2x{{y}^{2}}} \right)\left( {5z} \right)+{{{\left( {5z} \right)}}^{2}}} \right]\\&=\left( {2x{{y}^{2}}-5z} \right)\left( {4{{x}^{2}}{{y}^{4}}+10x{{y}^{2}}z+25{{z}^{2}}} \right)\end{align}$

Sum of Cubes

Difference of Cubes

$ 27{{y}^{3}}+1$

$ \begin{align}(a&=\sqrt[3]{{27{{y}^{3}}}}=3y,\,\,\,\,\,\,b=\sqrt[3]{{{{1}^{3}}}}=1)\\&=\left( {3y+1} \right)\left[ {{{{\left( {3y} \right)}}^{2}}-\left( {3y} \right)\left( 1 \right)+{{1}^{2}}} \right]\\&=\left( {3y+1} \right)\left( {9{{y}^{2}}-3y+1} \right)\end{align}$

$ 8{{x}^{3}}{{y}^{6}}-125{{z}^{3}}$

$ \begin{align}\,(a&=\sqrt[3]{{8{{x}^{3}}{{y}^{6}}}}=2x{{y}^{2}},\,\,\,\,\,\,b=\sqrt[3]{{125{{z}^{3}}}}=5z)\\&=\left( {2x{{y}^{2}}-5z} \right)\left[ {{{{\left( {2x{{y}^{2}}} \right)}}^{2}}+\left( {2x{{y}^{2}}} \right)\left( {5z} \right)+{{{\left( {5z} \right)}}^{2}}} \right]\\&=\left( {2x{{y}^{2}}-5z} \right)\left( {4{{x}^{2}}{{y}^{4}}+10x{{y}^{2}}z+25{{z}^{2}}} \right)\end{align}$

Factoring and Solving with Polynomials

Again, we first learned about factoring methods here in the Solving Quadratics by Factoring and Completing the Square section. Now that we know how to by Graphing and Finding Roots of Polynomial Functions section we can do fancier factoring, and thus find more roots. Remember that when we want to find solutions or roots, we set the equation to 0, factor, set each factor to 0 and solve. Here are some examples of factoring and solving polynomial equations; solve over the reals:

Equation	Solution(s)	Explanation
$ 2{{x}^{3}}-12{{x}^{2}}+18x=0$	$ \begin{array}{c}2{{x}^{3}}-12{{x}^{2}}+18x=0\\2x\left( {{{x}^{2}}-6x+9} \right)=0\\2x{{\left( {x-3} \right)}^{2}}=0\\2x=0;\,\,\,\,\,\,\,x-3=0\\x=0,\,\,\,3\end{array}$	First, take out the greatest common factor (GCF), which is $ 2x$. Then factor (“unFOIL”), but don’t forget to bring the “$ 2x$” down as part of the answer. (Remember when factoring that we need to find two numbers that add up to –6 but multiply to 9: –3 and –3.); this is actually a perfect square trinomial since when you double –3 we get –6, and when we square it, you get 9). Then set each factor to 0 to get the solutions or roots.
$ {{x}^{4}}+2{{x}^{3}}-27x=54$	$ \begin{array}{c}{{x}^{4}}+2{{x}^{3}}-27x=54\\{{x}^{4}}+2{{x}^{3}}-27x-54=0\\{{x}^{3}}\left( {x+2} \right)-27\left( {x+2} \right)=0\\\left( {{{x}^{3}}-27} \right)\left( {x+2} \right)=0\\{{x}^{3}}-27=0;\,\,\,x+2=0\\x=3,\,\,\,-2\end{array}$	In order to be able to solve this polynomial, first set it to 0, so subtract 54 from both sides. Then separate the four terms into two sets of two terms (grouping) to see if we can see a pattern when we take out the GCF from each set of term. (We do: we are left with $ x+20$!) Then, because of the (inverse of the) distributive property, take out the expression that’s repeated, use that as one factor, and use the coefficients of this expression as the other factor. Then set each factor to 0 to get the solutions or roots. Note that we know that the cube root of 27 is 3.
$ 16{{x}^{4}}-81=0$	$ \displaystyle \begin{array}{c}16{{x}^{4}}-81=0\\\left( {4{{x}^{4}}-9} \right)\left( {4{{x}^{4}}+9} \right)=0\\\left( {2{{x}^{2}}-3} \right)\left( {2{{x}^{2}}+3} \right)\left( {4{{x}^{4}}+9} \right)=0\\2{{x}^{2}}-3=0\end{array}$ $ \displaystyle x=\pm \frac{3}{2}$	We can see that we have a difference of squares, and then a difference of squares again. When we take the square root of a number, we have to consider both the positive and negative roots. Note that $ 2{{x}^{2}}+3$ and $ 4{{x}^{4}}+9$ (sum of squares) are prime, meaning we can’t factor them any further. Set the remaining factor to 0 to get the solutions or roots.
$ {{x}^{2}}-2\sqrt{6}x=-6$	$ \begin{array}{c}{{x}^{2}}-2\sqrt{6}x=-6\\{{x}^{2}}-2\sqrt{6}x+6=0\\{{\left( {x-\sqrt{6}} \right)}^{2}}=0\\x-\sqrt{6}=0\\x=\sqrt{6}\end{array}$	First make sure we set the equation to 0 be adding 6 to both sides. This one’s a little tricky, since we have a square root sign. But notice that if we halve the coefficient of the middle term and then square it, we get the last term! So we actually have a perfect square trinomial. Set the factor to 0 and solve for $ x$. Note that this could have been solved more easily using the Quadratic Formula.
$ {{x}^{3}}-64=0$	$ \begin{array}{c}\left( {x-4} \right)\left( {{{x}^{2}}+4x+16} \right)=0\\x-4=0\\x=4\end{array}$	There are actually two ways to do this; the easiest is to add 64 to both sides and then take the cube root of both sides to get $ x=4$. Factor the difference of cubes for practice; remember that: $ \begin{array}{l}{{a}^{3}}+{{b}^{3}}=\left( {a+b} \right)\left( {{{a}^{2}}-ab+{{b}^{2}}} \right)\\{{a}^{3}}-{{b}^{3}}=\left( {a-b} \right)\left( {{{a}^{2}}+ab+{{b}^{2}}} \right)\end{array}$. Since $ {{x}^{2}}+4x+16$ is prime, we only have $ x=4$ (by setting the factor to 0).

Here’s even more advanced solving, using techniques we will learn here in the Graphing and Finding Roots of Polynomial Functions section. Solve over the real and complex numbers:

Equation	Solution(s)	Explanation
$ 2{{x}^{3}}+3{{x}^{2}}-11x-6=0$	$ \displaystyle \begin{array}{c}2{{x}^{3}}+3{{x}^{2}}-11x-6=0\\\left( {x+3} \right)\left( {2{{x}^{2}}-3x-2} \right)=0\\\left( {x+3} \right)\left( {2x+1} \right)\left( {x-2} \right)=0\\x+3=0;\,\,\,2x+1=0;\,\,x-2=0\end{array}$ $ \displaystyle x=-3,\,\,-\frac{1}{2},\,\,2$	We can either use a graphing calculator, or use guess and check to find a first root in this polynomial. We know from the Rational Root Test that possible factors are: $ \displaystyle \require {cancel} \frac{{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}}{{\pm 1,\,\pm 2}}\,=\,\pm \,\,1,\,\pm \,\,2,\,\pm \,\,3,\,\pm \,\,6,\,\pm \,\,\frac{1}{2},\cancel{{\pm \,\,1}},\pm \,\,\frac{3}{2},\cancel{{\pm \,\,3}}$. –3 works (so we know that is a factor). Use synthetic division to get the remaining factor(s): \begin{array}{l}\left. {\underline {\, {\,-3\,\,} \,}}\! \right\| \,\,\,\,\,2\,\,\,\,\,\,\,\,3\,\,\,-11\,-6\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,6\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,-3\,\,\,-2\,\,\,\,\,\,\left\| \!{\overline {\, {\,\,0\,\,} \,}} \right. \end{array} We end up with $ 2{{x}^{2}}-3x-2$. Factor this to $ \left( {2x+1} \right)\left( {x-2} \right)$. Then set each factor to 0 to get the roots.
$ {{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}+5x=0$	$ \displaystyle \begin{array}{c}{{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}+5x=0\\x\left( {{{x}^{3}}+6{{x}^{2}}+6x+5} \right)=0\\x\left( {x+5} \right)\left( {{{x}^{2}}+x+1} \right)=0\\x=0;\,\,x+5=0;\,\,{{x}^{2}}+x+1=0\end{array}$ $ \displaystyle x=0,\,-5,\,-\frac{1}{2}-\frac{{\sqrt{3}\,}}{2}i,\,-\frac{1}{2}+\frac{{\sqrt{3}\,}}{2}i$	First factor out $ x$, which is the GCF. Then “guess” that $ -5$ is a root, using the Rational Root Test. Use synthetic division to find the remaining factor(s): \begin{array}{l}\left. {\underline {\, {\,-5\,\,} \,}}\! \right\| \,\,\,\,\,\,\,1\,\,\,\,\,\,6\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,5\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-5\,\,-5\,\,-5\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\,\,\,\,\left\| \!{\overline {\, {\,\,0\,\,} \,}} \right. \end{array} We end up with $ {{x}^{2}}+x+1$. It looks like we can’t factor, so use the Quadratic Formula and end up with complex roots: $ \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-1\pm \sqrt{{1-4\left( 1 \right)\left( 1 \right)}}}}{2}\\&=\frac{{-1\pm \sqrt{{-3}}}}{2}=\frac{{-1\pm \sqrt{3}\,i}}{2}\end{align}$. Set the other factors to 0 for all the roots.

Equation

Solution(s)

Explanation

$ 2{{x}^{3}}+3{{x}^{2}}-11x-6=0$

$ \displaystyle \begin{array}{c}2{{x}^{3}}+3{{x}^{2}}-11x-6=0\\\left( {x+3} \right)\left( {2{{x}^{2}}-3x-2} \right)=0\\\left( {x+3} \right)\left( {2x+1} \right)\left( {x-2} \right)=0\\x+3=0;\,\,\,2x+1=0;\,\,x-2=0\end{array}$

$ \displaystyle x=-3,\,\,-\frac{1}{2},\,\,2$

We can either use a graphing calculator, or use guess and check to find a first root in this polynomial. We know from the Rational Root Test that possible factors are: $ \displaystyle \require {cancel} \frac{{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}}{{\pm 1,\,\pm 2}}\,=\,\pm \,\,1,\,\pm \,\,2,\,\pm \,\,3,\,\pm \,\,6,\,\pm \,\,\frac{1}{2},\cancel{{\pm \,\,1}},\pm \,\,\frac{3}{2},\cancel{{\pm \,\,3}}$.

–3 works (so we know that is a factor). Use synthetic division to get the remaining factor(s):

\begin{array}{l}\left. {\underline {\,
{\,-3\,\,} \,}}\! \right| \,\,\,\,\,2\,\,\,\,\,\,\,\,3\,\,\,-11\,-6\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,6\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,-3\,\,\,-2\,\,\,\,\,\,\left| \!{\overline {\,
{\,\,0\,\,} \,}} \right. \end{array}

We end up with $ 2{{x}^{2}}-3x-2$. Factor this to $ \left( {2x+1} \right)\left( {x-2} \right)$. Then set each factor to 0 to get the roots.

$ {{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}+5x=0$

$ \displaystyle \begin{array}{c}{{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}+5x=0\\x\left( {{{x}^{3}}+6{{x}^{2}}+6x+5} \right)=0\\x\left( {x+5} \right)\left( {{{x}^{2}}+x+1} \right)=0\\x=0;\,\,x+5=0;\,\,{{x}^{2}}+x+1=0\end{array}$

$ \displaystyle x=0,\,-5,\,-\frac{1}{2}-\frac{{\sqrt{3}\,}}{2}i,\,-\frac{1}{2}+\frac{{\sqrt{3}\,}}{2}i$

First factor out $ x$, which is the GCF. Then “guess” that $ -5$ is a root, using the Rational Root Test. Use synthetic division to find the remaining factor(s):

\begin{array}{l}\left. {\underline {\,
{\,-5\,\,} \,}}\! \right| \,\,\,\,\,\,\,1\,\,\,\,\,\,6\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,5\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-5\,\,-5\,\,-5\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\,\,\,\,\left| \!{\overline {\,
{\,\,0\,\,} \,}} \right. \end{array}

We end up with $ {{x}^{2}}+x+1$. It looks like we can’t factor, so use the Quadratic Formula and end up with complex roots: $ \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-1\pm \sqrt{{1-4\left( 1 \right)\left( 1 \right)}}}}{2}\\&=\frac{{-1\pm \sqrt{{-3}}}}{2}=\frac{{-1\pm \sqrt{3}\,i}}{2}\end{align}$. Set the other factors to 0 for all the roots.

Factoring and Solving with Exponents

Factoring and Solving with Exponential Functions can be a bit trickier. Note that we learned about the properties of exponents here in the Exponents and Radicals in Algebra section, and did some solving with exponents here.

In your Pre-Calculus and Calculus classes, you may see algebraic exponential expressions that need factoring and possibly solving, either by taking out a Greatest Common Factor (GCF) or by “unFOILing”. These really aren’t that bad, if you remember a few hints:

To take out a GFC with exponents, take out the factor with the smallest exponent, whether it’s positive or negative. (Remember that “larger” negative numbers are actually smaller). This is even true for fractional exponents. Then, to get what’s left in the parentheses after you take out the GCF, subtract the exponents from the one you took out. For example, for the expression $ {{x}^{-5}}+{{x}^{-2}}+x$, take out $ {{x}^{-5}}$ for the GCF to get $ {{x}^{-5}}\left( {{x}^{-5-\,-5}}+{{x}^{-2-\,-5}}+{{x}^{1-\,-5}} \right)={{x}^{-5}}\left( {{x}^{0}}+{{x}^{3}}+{{x}^{6}} \right)={{x}^{-5}}\left( 1+{{x}^{3}}+{{x}^{6}} \right)$. (Remember that “ – – ” is the same as “+ +” or “+”). Multiply back to make sure you’ve factored correctly!
For fractional coefficients, find the common denominator, and take out the fraction that goes into all the other fractions. For the fraction you take out, the denominator is the least common denominator (LCD) of all the fractions, and the numerator is the Greatest Common Factor (GCF) of the numerators. For example, $ \displaystyle \frac{3}{4}{{x}^{2}}-\frac{1}{2}x+4=\frac{3}{4}{{x}^{2}}-\frac{2}{4}x+\frac{{16}}{4}=\frac{1}{4}\left( {3{{x}^{2}}-2x+16} \right)$ (since nothing except for 1 goes into 3 and 2 and 16). Multiply back to make sure you’ve factored correctly!
For a trinomial with a constant, if the largest exponent is twice that of the middle exponent, then use substitution like $ u$, for the middle exponent, “unFOIL”, and then put the “real” expression back in. For example, for $ {{x}^{\frac{2}{3}}}-{{x}^{\frac{1}{3}}}-2$ , let $ u={{x}^{\frac{1}{3}}}$ , and we have $ {{u}^{2}}-u-2$, which factors to $ \left( u-2 \right)\left( u+1 \right)$. Then substitute back to $ \left( {{x}^{\frac{1}{3}}}-2 \right)\left( {{x}^{\frac{1}{3}}}+1 \right)$ and solve from there (set each to 0 and solve). Always multiply back to make sure you’ve factored correctly. We call this method u-substitution or simply u-sub.

Let’s do some factoring. Learning to factor these will actually help you a lot when you get to Calculus:

Math	Notes
$ \displaystyle 4{{\left( {x-2} \right)}^{{-9}}}-2{{\left( {x-2} \right)}^{{-10}}}$ $ \displaystyle \begin{align}&=2{{\left( {x-2} \right)}^{{-10}}}\left[ {2{{{\left( {x-2} \right)}}^{{-9-\left( {-10} \right)}}}-{{{\left( {x-2} \right)}}^{{-10-\left( {-10} \right)}}}} \right]\\&=2{{\left( {x-2} \right)}^{{-10}}}\left[ {2{{{\left( {x-2} \right)}}^{1}}-{{{\left( {x-2} \right)}}^{0}}} \right]\\&=2{{\left( {x-2} \right)}^{{-10}}}\left( {2x-4-1} \right)\\&=2{{\left( {x-2} \right)}^{{-10}}}\left( {2x-5} \right)=\frac{{2\left( {2x-5} \right)}}{{{{{\left( {x-2} \right)}}^{{10}}}}}\end{align}$	Take out the common factor (GCF) of the coefficients, and the smallest exponent of the algebraic terms: $ 2{{\left( {x-2} \right)}^{{-10}}}$. Since the opposite of multiplication is division, subtract exponents when dividing, to get the other factors. (Try this with easier exponents to convince yourself!). When subtracting exponents, be sure to watch signs (remember that two negatives in a row is a positive). Also remember than anything raised to 0 is just 1. Then combine terms. It’s a good idea to multiply back before you combine terms to make sure you took out the GCF correctly!
$ {{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}+5{{\left( {{{x}^{2}}-4} \right)}^{{\frac{2}{5}}}}$ $ \begin{align}&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left[ {{{{\left( {{{x}^{2}}-4} \right)}}^{{-\frac{3}{5}-\left( {-\frac{3}{5}} \right)}}}+5{{{\left( {{{x}^{2}}-4} \right)}}^{{\frac{2}{5}-\left( {-\frac{3}{5}} \right)}}}} \right]\\&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left[ {{{{\left( {{{x}^{2}}-4} \right)}}^{0}}+5{{{\left( {{{x}^{2}}-4} \right)}}^{1}}} \right]\\&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left( {1+\,\,\,\,5{{x}^{2}}-20} \right)\\&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left( {5{{x}^{2}}-19} \right)=\frac{{5{{x}^{2}}-19}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{{\frac{3}{5}}}}}}=\frac{{5{{x}^{2}}-19}}{{\sqrt[5]{{{{{\left( {{{x}^{2}}-4} \right)}}^{3}}}}}}\end{align}$	We do this one the same way, but be careful with the fractional exponents. Again, take out the algebraic terms with the smallest exponent: $ {{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}$. Note that $ \displaystyle \frac{2}{5}-\left( {-\frac{3}{5}} \right)=\frac{5}{5}=1$.
$ \displaystyle 3{{x}^{{\frac{9}{2}}}}-15{{x}^{{\frac{7}{2}}}}+12{{x}^{{\frac{5}{2}}}}$ $ \displaystyle \begin{align}&=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{{\frac{9}{2}-\frac{5}{2}}}}-5{{x}^{{\frac{7}{2}-\frac{5}{2}}}}+4{{x}^{{\frac{5}{2}-\frac{5}{2}}}}} \right)\\&=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{2}}-5{{x}^{1}}+4{{x}^{0}}} \right)\\&=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{2}}-5x+4} \right)=3{{x}^{{\frac{5}{2}}}}\left( {x-4} \right)\left( {x-1} \right)\end{align}$	This is a trinomial that we might be able to factor (“unFOIL”), but we still need to take out any possible GCF first. Take out the GCF of the coefficients, and then the exponential term with the smallest exponent. Note then that we can “unFOIL” after we take out the GCF.
$ \displaystyle \frac{3}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}+\frac{5}{4}{{\left( {x-5} \right)}^{{\frac{1}{4}}}}$ $ \displaystyle \begin{align}&=\frac{3}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}+\frac{{10}}{8}{{\left( {x-5} \right)}^{{\frac{1}{4}}}}\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {3{{{\left( {x-5} \right)}}^{{-\frac{3}{4}-\left( {-\frac{3}{4}} \right)}}}+10{{{\left( {x-5} \right)}}^{{\frac{1}{4}-\left( {-\frac{3}{4}} \right)}}}} \right]\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {3{{{\left( {x-5} \right)}}^{0}}+10{{{\left( {x-5} \right)}}^{1}}} \right]\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {3+10\left( {x-5} \right)} \right]\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {10x-47} \right]=\frac{{10x-47}}{{8{{{\left( {x-5} \right)}}^{{\frac{3}{4}}}}}}=\frac{{10x-47}}{{8\sqrt[4]{{{{{\left( {x-5} \right)}}^{3}}}}}}\end{align}$	Find the common denominator (8), and take out the fraction that goes into all the other fractions $ \displaystyle \left( {\frac{1}{8}} \right)$. Again, for the fraction you take out, the denominator is the LCD of all the fractions (8), and the numerator is the GCF of the numerators (only 1 goes into both 3 and 10). Notice that when you take the $ \displaystyle \frac{1}{8}$ out, you are just left with the numerators (3 and 10, respectively). Always multiply back (before combining terms) to make sure you took out the GCF correctly.

Math

Notes

$ \displaystyle 4{{\left( {x-2} \right)}^{{-9}}}-2{{\left( {x-2} \right)}^{{-10}}}$

$ \displaystyle \begin{align}&=2{{\left( {x-2} \right)}^{{-10}}}\left[ {2{{{\left( {x-2} \right)}}^{{-9-\left( {-10} \right)}}}-{{{\left( {x-2} \right)}}^{{-10-\left( {-10} \right)}}}} \right]\\&=2{{\left( {x-2} \right)}^{{-10}}}\left[ {2{{{\left( {x-2} \right)}}^{1}}-{{{\left( {x-2} \right)}}^{0}}} \right]\\&=2{{\left( {x-2} \right)}^{{-10}}}\left( {2x-4-1} \right)\\&=2{{\left( {x-2} \right)}^{{-10}}}\left( {2x-5} \right)=\frac{{2\left( {2x-5} \right)}}{{{{{\left( {x-2} \right)}}^{{10}}}}}\end{align}$

Take out the common factor (GCF) of the coefficients, and the smallest exponent of the algebraic terms: $ 2{{\left( {x-2} \right)}^{{-10}}}$.

Since the opposite of multiplication is division, subtract exponents when dividing, to get the other factors. (Try this with easier exponents to convince yourself!). When subtracting exponents, be sure to watch signs (remember that two negatives in a row is a positive). Also remember than anything raised to 0 is just 1. Then combine terms. It’s a good idea to multiply back before you combine terms to make sure you took out the GCF correctly!

$ {{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}+5{{\left( {{{x}^{2}}-4} \right)}^{{\frac{2}{5}}}}$

$ \begin{align}&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left[ {{{{\left( {{{x}^{2}}-4} \right)}}^{{-\frac{3}{5}-\left( {-\frac{3}{5}} \right)}}}+5{{{\left( {{{x}^{2}}-4} \right)}}^{{\frac{2}{5}-\left( {-\frac{3}{5}} \right)}}}} \right]\\&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left[ {{{{\left( {{{x}^{2}}-4} \right)}}^{0}}+5{{{\left( {{{x}^{2}}-4} \right)}}^{1}}} \right]\\&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left( {1+\,\,\,\,5{{x}^{2}}-20} \right)\\&={{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}\left( {5{{x}^{2}}-19} \right)=\frac{{5{{x}^{2}}-19}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{{\frac{3}{5}}}}}}=\frac{{5{{x}^{2}}-19}}{{\sqrt[5]{{{{{\left( {{{x}^{2}}-4} \right)}}^{3}}}}}}\end{align}$

We do this one the same way, but be careful with the fractional exponents. Again, take out the algebraic terms with the smallest exponent: $ {{\left( {{{x}^{2}}-4} \right)}^{{-\frac{3}{5}}}}$.

Note that $ \displaystyle \frac{2}{5}-\left( {-\frac{3}{5}} \right)=\frac{5}{5}=1$.

$ \displaystyle 3{{x}^{{\frac{9}{2}}}}-15{{x}^{{\frac{7}{2}}}}+12{{x}^{{\frac{5}{2}}}}$

$ \displaystyle \begin{align}&=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{{\frac{9}{2}-\frac{5}{2}}}}-5{{x}^{{\frac{7}{2}-\frac{5}{2}}}}+4{{x}^{{\frac{5}{2}-\frac{5}{2}}}}} \right)\\&=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{2}}-5{{x}^{1}}+4{{x}^{0}}} \right)\\&=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{2}}-5x+4} \right)=3{{x}^{{\frac{5}{2}}}}\left( {x-4} \right)\left( {x-1} \right)\end{align}$

This is a trinomial that we might be able to factor (“unFOIL”), but we still need to take out any possible GCF first.

Take out the GCF of the coefficients, and then the exponential term with the smallest exponent.

Note then that we can “unFOIL” after we take out the GCF.

$ \displaystyle \frac{3}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}+\frac{5}{4}{{\left( {x-5} \right)}^{{\frac{1}{4}}}}$

$ \displaystyle \begin{align}&=\frac{3}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}+\frac{{10}}{8}{{\left( {x-5} \right)}^{{\frac{1}{4}}}}\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {3{{{\left( {x-5} \right)}}^{{-\frac{3}{4}-\left( {-\frac{3}{4}} \right)}}}+10{{{\left( {x-5} \right)}}^{{\frac{1}{4}-\left( {-\frac{3}{4}} \right)}}}} \right]\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {3{{{\left( {x-5} \right)}}^{0}}+10{{{\left( {x-5} \right)}}^{1}}} \right]\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {3+10\left( {x-5} \right)} \right]\\&=\frac{1}{8}{{\left( {x-5} \right)}^{{-\frac{3}{4}}}}\left[ {10x-47} \right]=\frac{{10x-47}}{{8{{{\left( {x-5} \right)}}^{{\frac{3}{4}}}}}}=\frac{{10x-47}}{{8\sqrt[4]{{{{{\left( {x-5} \right)}}^{3}}}}}}\end{align}$

Find the common denominator (8), and take out the fraction that goes into all the other fractions $ \displaystyle \left( {\frac{1}{8}} \right)$. Again, for the fraction you take out, the denominator is the LCD of all the fractions (8), and the numerator is the GCF of the numerators (only 1 goes into both 3 and 10). Notice that when you take the $ \displaystyle \frac{1}{8}$ out, you are just left with the numerators (3 and 10, respectively).

Always multiply back (before combining terms) to make sure you took out the GCF correctly.

After factoring, you may be asked to solve the exponential equation. Here are some examples, some using u-substitution. Note that the third problem uses log solving from here in the Logarithmic Functions section.

Math	Notes
$ \displaystyle 2{{x}^{{\frac{1}{2}}}}+{{x}^{{\frac{1}{4}}}}-15=0$ Let $ \displaystyle u={{x}^{{\frac{1}{4}}}}$ $ \displaystyle \begin{array}{c}2{{u}^{2}}+u-15=0\,\,\,\,\,\,\,\,\,\,\left( {2u-5} \right)\left( {u+3} \right)=0\end{array}$ $ \displaystyle u=\frac{5}{2},-3$ Substitute $ \displaystyle {{x}^{{\frac{1}{4}}}}$ back in for $ u$ for both factors: $ \displaystyle {{x}^{{\frac{1}{4}}}}=\frac{5}{2};\,\,\,{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{4}}={{\left( {\frac{5}{2}} \right)}^{4}};\,\,\,x=\frac{{625}}{{16}}$ $ \require{cancel} \displaystyle \cancel{{{{x}^{{\frac{1}{4}}}}=-3}}$; No Solution Check:	Use $ u$-substitution here, since the largest exponent $ \left( {\frac{1}{2}} \right)$ is twice that of the middle exponent $ \left( {\frac{1}{4}} \right)$. Use the trick to let $ u$ equal to the variable in the middle: $ u={{x}^{{\frac{1}{4}}}}$, and then $ \displaystyle 2{{x}^{{\frac{1}{2}}}}=2{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{2}}=2{{u}^{2}}$. (Remember that when we raise an exponent to another exponent, we multiply exponents). Now we can factor more easily. We can factor this one; if you can’t, you can always use the Quadratic Formula. After factoring, put the $ u$’s back in. We want to end up with $ \displaystyle {{x}^{1}}$ by itself, so raise both sides to the reciprocal of the exponent. For the second factor, we have no solution, since we have an even exponent in the denominator (which is an even root) and we need to end up with a negative number; this can’t be done with real numbers. Try the answer back in for $ x$ to make sure it works; $ \displaystyle \frac{{625}}{{16}}$ works! You can use the STO> X function in your graphing calculator to check (see left).
$ \displaystyle {{9}^{x}}+{{3}^{{x+1}}}=4$ $ \displaystyle \begin{align}{{\left( {{{3}^{2}}} \right)}^{x}}+\left( {{{3}^{x}}} \right)\left( {{{3}^{1}}} \right)&=4\\{{3}^{{2x}}}+\left( 3 \right)\left( {{{3}^{x}}} \right)&=4\\{{\left( {{{3}^{x}}} \right)}^{2}}+\left( 3 \right)\left( {{{3}^{x}}} \right)-4&=0\end{align}$ Let $ \displaystyle u={{3}^{x}}$ $ \begin{array}{c}{{u}^{2}}+3u-4=0\,\,\,\,\,\,\,\,\,\,\left( {u-1} \right)\left( {u+4} \right)=0\\u=1,\,-4\end{array}$ Substitute $ {{3}^{x}}$ back in for $ u$ for both factors: $ {{3}^{x}}=1;\,\,\,\,\,x=0$ $ \cancel{{{{3}^{x}}=-4}}$; No Solution	We need the “3”’s to match, so break up the second term and change the base of the first term – it works! Remember that when we multiply the same base, we add exponents. Also remember that when we raise an exponent to another exponent, we multiply exponents. Let $ u={{3}^{x}}$, and then $ {{9}^{x}}={{\left( {{{3}^{2}}} \right)}^{x}}={{3}^{{2x}}}={{3}^{{x\cdot 2}}}={{\left( {{{3}^{x}}} \right)}^{2}}={{u}^{2}}$. For the first factor, we know that anything raised to 0 = 1, so the solution is 0. For the second factor, we get no solution, since we can’t raise 3 to anything to get a negative number. Try the 0 back in the original and see if it works – it does: $ {{9}^{0}}+\left( 3 \right){{3}^{0}}=4;\,\,\,\,1+3\left( 1 \right)=4$
$ {{\left( {{{3}^{x}}-27} \right)}^{3}}+18{{\left( {{{3}^{x}}-27} \right)}^{2}}=0$ $ \displaystyle \begin{array}{c}{{\left( {{{3}^{x}}-27} \right)}^{2}}\left[ {{{{\left( {{{3}^{x}}-27} \right)}}^{1}}+18} \right]=0\\{{\left( {{{3}^{x}}-27} \right)}^{2}}\left( {{{3}^{x}}-9} \right)=0\\\sqrt{{{{{\left( {{{3}^{x}}-27} \right)}}^{2}}}}=\pm 0=0;\,\,\,\,\,\,\,\left( {{{3}^{x}}-9} \right)=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}^{x}}=27;\,\,\,\,\,\,\,\,\,{{3}^{x}}=9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=3;\,\,\,\,\,\,\,\,\,x=2\end{array}$	For this one, we don’t really need $ u$-sub, but since we can just take out the expression with the smallest exponent, like we did earlier. After factoring, set each factor to 0. Think about what exponents will work when 3 is raised to that exponent – to get 27 and 9 (or we can used logs). Both 3 and 2 work; plug them back in the original to make sure they are correct.
Here’s one where we have to use a Logarithmic Function to solve: $ 2{{e}^{{5x}}}-5{{e}^{{2x}}}-12{{e}^{{-x}}}=0$ $ \displaystyle {{e}^{{-x}}}\left( {2{{e}^{{6x}}}-5{{e}^{{3x}}}-12} \right)=0$ $ \displaystyle \begin{array}{c}\text{Let }u={{e}^{{3x}}}\\{{e}^{{-x}}}\left( {2{{u}^{2}}-5u-12} \right)=0\\{{e}^{{-x}}}\left( {2u+3} \right)\left( {u-4} \right)=0\end{array}$ $ \displaystyle \begin{align}{{e}^{{-x}}}=0;\,\,\,\,u=-\frac{3}{2};\,\,\,\,u=4\,\,\,\,\\{{e}^{{-x}}}=0;\,\,\,{{e}^{{3x}}}=-\frac{3}{2};\,\,\,{{e}^{{3x}}}=4\,\,\,\,\,\,\,\,\\\,\,\,\,\cancel{{{{e}^{{-x}}}=0}};\,\,\,\cancel{{{{e}^{{3x}}}=-\frac{3}{2}}};\,\,\,\,\ln {{e}^{{3x}}}=\ln 4\,\,\,\,\,\,\,\,\\3x=\ln 4;\,\,\,\,x=\frac{{\ln 4}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\end{align}$	First, factor out the GCF, which is $ {{3}^{{-x}}}$ (smallest exponent). Then, set $ u={{e}^{{3x}}}$, since $ \displaystyle {{e}^{{6x}}}={{\left( {{{e}^{{3x}}}} \right)}^{2}}$ (note that $ u$ is in the middle term). Factor, and set each factor to 0 (including the GCF). Note that we had to “throw away” the extraneous solutions, since (positive) $ e$ raised to anything can’t be 0 or negative.

Math

Notes

$ \displaystyle 2{{x}^{{\frac{1}{2}}}}+{{x}^{{\frac{1}{4}}}}-15=0$

Let $ \displaystyle u={{x}^{{\frac{1}{4}}}}$

$ \displaystyle \begin{array}{c}2{{u}^{2}}+u-15=0\,\,\,\,\,\,\,\,\,\,\left( {2u-5} \right)\left( {u+3} \right)=0\end{array}$

$ \displaystyle u=\frac{5}{2},-3$

Substitute $ \displaystyle {{x}^{{\frac{1}{4}}}}$ back in for $ u$ for both factors:

$ \displaystyle {{x}^{{\frac{1}{4}}}}=\frac{5}{2};\,\,\,{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{4}}={{\left( {\frac{5}{2}} \right)}^{4}};\,\,\,x=\frac{{625}}{{16}}$

$ \require{cancel} \displaystyle \cancel{{{{x}^{{\frac{1}{4}}}}=-3}}$; No Solution

Check:

Use $ u$-substitution here, since the largest exponent $ \left( {\frac{1}{2}} \right)$ is twice that of the middle exponent $ \left( {\frac{1}{4}} \right)$. Use the trick to let $ u$ equal to the variable in the middle: $ u={{x}^{{\frac{1}{4}}}}$, and then $ \displaystyle 2{{x}^{{\frac{1}{2}}}}=2{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{2}}=2{{u}^{2}}$. (Remember that when we raise an exponent to another exponent, we multiply exponents). Now we can factor more easily. We can factor this one; if you can’t, you can always use the Quadratic Formula.

After factoring, put the $ u$’s back in. We want to end up with $ \displaystyle {{x}^{1}}$ by itself, so raise both sides to the reciprocal of the exponent. For the second factor, we have no solution, since we have an even exponent in the denominator (which is an even root) and we need to end up with a negative number; this can’t be done with real numbers.

Try the answer back in for $ x$ to make sure it works; $ \displaystyle \frac{{625}}{{16}}$ works! You can use the STO> X function in your graphing calculator to check (see left).

$ \displaystyle {{9}^{x}}+{{3}^{{x+1}}}=4$

$ \displaystyle \begin{align}{{\left( {{{3}^{2}}} \right)}^{x}}+\left( {{{3}^{x}}} \right)\left( {{{3}^{1}}} \right)&=4\\{{3}^{{2x}}}+\left( 3 \right)\left( {{{3}^{x}}} \right)&=4\\{{\left( {{{3}^{x}}} \right)}^{2}}+\left( 3 \right)\left( {{{3}^{x}}} \right)-4&=0\end{align}$

Let $ \displaystyle u={{3}^{x}}$

$ \begin{array}{c}{{u}^{2}}+3u-4=0\,\,\,\,\,\,\,\,\,\,\left( {u-1} \right)\left( {u+4} \right)=0\\u=1,\,-4\end{array}$

Substitute $ {{3}^{x}}$ back in for $ u$ for both factors: $ {{3}^{x}}=1;\,\,\,\,\,x=0$

$ \cancel{{{{3}^{x}}=-4}}$; No Solution

We need the “3”’s to match, so break up the second term and change the base of the first term – it works! Remember that when we multiply the same base, we add exponents. Also remember that when we raise an exponent to another exponent, we multiply exponents.

Let $ u={{3}^{x}}$, and then $ {{9}^{x}}={{\left( {{{3}^{2}}} \right)}^{x}}={{3}^{{2x}}}={{3}^{{x\cdot 2}}}={{\left( {{{3}^{x}}} \right)}^{2}}={{u}^{2}}$.

For the first factor, we know that anything raised to 0 = 1, so the solution is 0. For the second factor, we get no solution, since we can’t raise 3 to anything to get a negative number. Try the 0 back in the original and see if it works – it does:

$ {{9}^{0}}+\left( 3 \right){{3}^{0}}=4;\,\,\,\,1+3\left( 1 \right)=4$

$ {{\left( {{{3}^{x}}-27} \right)}^{3}}+18{{\left( {{{3}^{x}}-27} \right)}^{2}}=0$

$ \displaystyle \begin{array}{c}{{\left( {{{3}^{x}}-27} \right)}^{2}}\left[ {{{{\left( {{{3}^{x}}-27} \right)}}^{1}}+18} \right]=0\\{{\left( {{{3}^{x}}-27} \right)}^{2}}\left( {{{3}^{x}}-9} \right)=0\\\sqrt{{{{{\left( {{{3}^{x}}-27} \right)}}^{2}}}}=\pm 0=0;\,\,\,\,\,\,\,\left( {{{3}^{x}}-9} \right)=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}^{x}}=27;\,\,\,\,\,\,\,\,\,{{3}^{x}}=9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=3;\,\,\,\,\,\,\,\,\,x=2\end{array}$

For this one, we don’t really need $ u$-sub, but since we can just take out the expression with the smallest exponent, like we did earlier.

After factoring, set each factor to 0. Think about what exponents will work when 3 is raised to that exponent – to get 27 and 9 (or we can used logs). Both 3 and 2 work; plug them back in the original to make sure they are correct.

Here’s one where we have to use a Logarithmic Function to solve:

$ 2{{e}^{{5x}}}-5{{e}^{{2x}}}-12{{e}^{{-x}}}=0$

$ \displaystyle {{e}^{{-x}}}\left( {2{{e}^{{6x}}}-5{{e}^{{3x}}}-12} \right)=0$

$ \displaystyle \begin{array}{c}\text{Let }u={{e}^{{3x}}}\\{{e}^{{-x}}}\left( {2{{u}^{2}}-5u-12} \right)=0\\{{e}^{{-x}}}\left( {2u+3} \right)\left( {u-4} \right)=0\end{array}$

$ \displaystyle \begin{align}{{e}^{{-x}}}=0;\,\,\,\,u=-\frac{3}{2};\,\,\,\,u=4\,\,\,\,\\{{e}^{{-x}}}=0;\,\,\,{{e}^{{3x}}}=-\frac{3}{2};\,\,\,{{e}^{{3x}}}=4\,\,\,\,\,\,\,\,\\\,\,\,\,\cancel{{{{e}^{{-x}}}=0}};\,\,\,\cancel{{{{e}^{{3x}}}=-\frac{3}{2}}};\,\,\,\,\ln {{e}^{{3x}}}=\ln 4\,\,\,\,\,\,\,\,\\3x=\ln 4;\,\,\,\,x=\frac{{\ln 4}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\end{align}$

First, factor out the GCF, which is $ {{3}^{{-x}}}$ (smallest exponent).

Then, set $ u={{e}^{{3x}}}$, since $ \displaystyle {{e}^{{6x}}}={{\left( {{{e}^{{3x}}}} \right)}^{2}}$ (note that $ u$ is in the middle term). Factor, and set each factor to 0 (including the GCF).

Note that we had to “throw away” the extraneous solutions, since (positive) $ e$ raised to anything can’t be 0 or negative.

Learn these rules, and practice, practice, practice!

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