# 1.1: Solve Polynomial Equations by Factoring

footfall 3 : look for factors that can be factored far. tone 2 : Determine the number of terms in the polynomial. step 1 : check for park factors. If the terms have common factors, then factor out the greatest common factor ( GCF ). We have learned assorted techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method acting to apply. The following outlines a general road map for factoring polynomials.

## Use Factoring to Solve Equations

We will first solve some equations by using the Zero Factor Property. The Zero Factor Property ( besides called the Zero Product Property ) says that if the product of two quantities is zero, then at least one of those quantities is zero. The alone way to get a product equal to zero is to multiply by zero itself .
Zero Factor Property
If \ ( a·b=0\ ), then either \ ( a=0\ ) or \ ( b=0\ ) or both .
For case, consider the equation \ ( ( x – 3 ) ( x – 2 ) = 0 \ ). According to the Zero Factor Property, this product can alone be zero if one of the factors is zero. For this equation, the factors are \ ( ( x-3 ) \ ) and \ ( ( x-2 ) \ ). Factors are the expressions that are multiplied in concert to form a \ ( \underline { \text { intersection } } \ ) .
\ ( ( x – 3 ) ( x – 2 ) = 0 \ )
\ ( \begin { align * } x-3 & =0 & \text { or } & & x-2 & =0 \\
x & =3 & & & x & =2
\end { align * } \ )
\ ( \text { Solution Set : } \ { 3, 2\ } \ )
These proposed solutions can be checked by substituting back in the original equation .
\ ( \begin { align * }
\text { Check } x & = { \color { Cerulean } { 3 } } & \text { and } & & \text { Check } x & = { \color { Cerulean } { 2 } } \\
( { \color { Cerulean } { 3 } } – 3 ) ( { \color { Cerulean } { 3 } } – 2 ) & =0 & & & ( { \color { Cerulean } { 2 } } – 3 ) ( { \color { Cerulean } { 2 } } – 2 ) & =0 \\
( 0 ) ( 1 ) & =0 \quad { \color { Cerulean } { ✓ } } & & & ( -1 ) ( 0 ) & =0 \quad { \color { Cerulean } { ✓ } }
\end { align * } \ )
How to : Use the Zero Factor Property to Solve an equality .

1. ZERO. Write the equation so one side of the equation is  zero. Write the expression on the other side of the equal sign in order of descending powers of $$x$$ with a positive coefficient on the term with highest exponent.
2. FACTORFactor the expression.
3. PROPERTY. Set each factor equal to zero and solve. (This is the property — a factor that is zero will make the product of factors it is a part of also equal to zero).  The solutions obtained are the values of $$x$$ that will make the original equation a true statement.
4. Check by substituting solutions into the original equation.

exercise \ ( \PageIndex { 1 } \ ) component out a GCF
clear : \ ( 2 x ^2 = 8x \ ) .
Solution.   Notice that the first gradation requires one slope of the equation to be made zero. If both sides of the equal polarity rather were first divided by ten, then only one solution \ ( x=4\ ) would have been found. Dividing by a varying expression can result in lose solutions !
\ ( \begin { align * }
& & 2x^2-8x & =0 & & & & \text { 1. } \underline { \textbf { Zero } } \text {. Make one side zero } \\
& & 2x ( x-4 ) & = 0 & & & & \text { 2. } \underline { \textbf { Factor } } \text {. gene out the GCF } \\
2x & =0 & \text { or } & & x-4 & =0 & & \text { 3. } \underline { \textbf { property } } \text {. Set each factor to 0 } \\
x & =0 & & & x & =4 & & \quad \text { And solve } \\
& & \text { Solution Set : } & \ { 0, 4 \ } \end { align * } \ )
model \ ( \PageIndex { 2 } \ ) Factor four terms by grouping pairs
solve : \ ( 4 x ^ { 3 } – ten ^ { 2 } – 100 adam + 25 = 0\ ) .
Solution
\ ( \begin { align * }
& & 4 ten ^ { 3 } – adam ^ { 2 } – 100 ten + 25 = 0 & & & & & \text { 1. } \underline { \textbf { Zero } } \text {. One side already zero } \\
& & x ^ { 2 } ( 4 x – 1 ) – 25 ( 4 x – 1 ) =0 & & & & & \text { 2. } \underline { \textbf { Factor } } \text {. factor by grouping pairs } \\
& & ( 4 x – 1 ) \left ( ten ^ { 2 } – 25 \right ) =0 & & & & & \quad\text { Factor out the coarse binomial } \\
& & ( 4 x – 1 ) ( x – 5 ) ( ten + 5 ) =0 & & & & & \quad\text { Factor a difference of squares } \\
4x-1 & =0 & x-5=0\quad\qquad & & x+5 & =0 & & \text { 3. } \underline { \textbf { property } } \text {. Set each component to 0 } \\
4x & =1 & x=5\qquad\qquad & & x & =-5 & & \quad \text { And solve } \\
x & =\frac { 1 } { 4 } & & & & & & \\
& & \text { Solution Set : } \Big\ { \frac { 1 } { 4 }, 5, -5 \Big\ } & \end { align * } \ )
example \ ( \PageIndex { 3 } \ ) agent a trinomial ( with a changeless GCF and then \ ( a = 1\ ) )
resolve : \ ( 3x^2=12x+63\ ) .
Solution.  This example  highlights the essential first step of making one side of the equation zero before the Zero Factor Property is applied.  Also, factoring produces three factors, but the first factor is a constant $$(3)$$ which can never be equal to $$0$$

This case highlights the essential first footfall of making one slope of the equality zero before the Zero Factor Property is applied. besides, factoring produces three factors, but the first factor is a constant \ ( ( 3 ) \ ) which can never be equal to \ ( 0 \ ) \ ( \begin { align * }
& & 3x^2−12x−63=0 & & & & & \text { 1. } \underline { \textbf { Zero } } \text {. Make one english zero } \\
& & 3 ( x^2−4x−21 ) =0 & & & & & \text { 2. } \underline { \textbf { Factor } } \text {. factor out the GCF } \\
& & 3 ( x−7 ) ( x+3 ) =0 & & & & & \quad\text { Factor the trinomial } \\
3 & =0 & x-7=0\quad & & x+3 & =0 & & \text { 3. } \underline { \textbf { property } } \text {. Set each factor to 0 } \\
3 & \neq 0 & x=7\qquad & & x & =-3 & & \quad \text { And solve } \\
& & \text { Solution Set : } \ { 7, -3 \ } & \end { align * } \ )
exercise \ ( \PageIndex { 4 } \ ) factor a trinomial ( with \ ( a \ne 1\ ) )
resolve : \ ( 15 x ^ { 2 } + 3 x – 8 = 5 x – 7\ ) .
Solution
\ ( \begin { array } { chlorine }
15 x ^ { 2 } + 3 x – 8 = 5 x – 7 & \\
15 ten ^ { 2 } – 2 x – 1 = 0 & \qquad \text { 1. } \underline { \textbf { Zero } } \text {. Make one side zero } \\
( 3x−1 ) ( 5x+1 ) =0 & \qquad \text { 2. } \underline { \textbf { Factor } } \text {. Factor the trinomial } \\
\begin { range } { three hundred } 3x−1=0 & \text { and } & 5x+1=0 \\
3x = 1 & & 5x = -1 \\
x = \frac { 1 } { 3 } & & x = -\frac { 1 } { 5 } \end { align }
& \begin { array } { lambert } \qquad \text { 3. } \underline { \textbf { property } } \text {. Set each gene to 0 } \\
\quad \qquad \text { And solve } \\ \text { } \end { range } \\
\text { Solution Set : } \Big\ { \frac { 1 } { 3 }, -\frac { 1 } { 5 } \Big\ }
\end { align } \ )
example \ ( \PageIndex { 5 } \ ) factor a trinomial ( with \ ( a \ne 1\ ) )

solve : \ ( ( 3x−8 ) ( x−1 ) =3x\ ) .
Solution.  This quadratic equation equation appears to be factored ; therefore it might be tempting to set each factor equal to \ ( 3x\ ). however, this would lead to incorrect results. We must inaugural rewrite the equation equal to zero, so that we can apply the zero-product property
\ ( \begin { array } { cccl }
& ( 3x−8 ) ( x−1 ) =3x & \\
& 3x^2−11x+8=3x & & \qquad \text { 1. } \underline { \textbf { Zero } } \text {. Make one side zero } \\
& 3x^2−14x+8=0 & & \\
& ( 3x−2 ) ( x−4 ) =0 & & \qquad \text { 2. } \underline { \textbf { Factor } } \text {. Factor the trinomial } \\
3x−2=0 & \text { and } & x-4=0 & \qquad \text { 3. } \underline { \textbf { property } } \text {. Set each divisor to 0 } \\
3x = 2 & & x=4 & \quad \qquad \text { And solve } \\
x = \frac { 2 } { 3 } & & \\
& \text { Solution Set : } \Big\ { \frac { 2 } { 3 }, 4 \Big\ } &
\end { array } \ )
exemplar \ ( \PageIndex { 6 } \ ) gene a difference of squares
solve : \ ( 169q^2=49\ ) .
Solution
\ ( \begin { array } { cccl }
& 169q^2=49 & \\
& 169x^2−49=0 & & \qquad \text { 1. } \underline { \textbf { Zero } } \text {. Make one side zero } \\
& ( 13x−7 ) ( 13x+7 ) =0 & & \qquad \text { 2. } \underline { \textbf { Factor } } \text {. component a remainder of squares } \\
13x−7=0 & \text { and } & 13x+7=0 & \qquad \text { 3. } \underline { \textbf { place } } \text {. Set each factor to 0 } \\
13x = 7 & & 13x=-7 & \quad \qquad \text { And solve } \\
x = \frac { 7 } { 13 } & & x = -\frac { 7 } { 13 } \\
& \text { Solution Set : } \Big\ { \frac { 7 } { 13 }, -\frac { 7 } { 13 } \Big\ } &
\end { array } \ )

exemplar \ ( \PageIndex { 7 } \ ) factor a perfective square trinomial ( with a variable star GCF )
solve : \ ( 9m^3+100m=60m^2\ )
Solution
\ ( \begin { array } { one hundred fifty }
9m^3+100m=60m^2 & \\
9m^3−60m^2+100m=0 & \text { 1. } \underline { \textbf { Zero } } \text {. Make one side zero } \\
thousand ( 9m^2−60m+100 ) =0 & \text { 2. } \underline { \textbf { Factor } } \text {. component out the GCF } \\
megabyte ( 3m−10 ) ^2=0 & \text { Factor the Perfect Square trinomial } \\
\begin { array } { three hundred }
m=0 & \text { and } & 3m-10=0 \\
& & m=10 \\
& & m=\frac { 10 } { 3 }
\end { range }
&
\begin { array } { lambert } \text { 3. } \underline { \textbf { property } } \text {. Set each factor to 0 } \\
\text { And solve } \\ \text { } \end { align }
\\
\text { Solution Set : } \Big\ { 0, \frac { 10 } { 3 } \Big\ } \end { array } \ )
case \ ( \PageIndex { 8 } \ ) Factor four terms ( the dispute of a binomial square and a monomial straight )
resolve : \ ( 4x^ { 2 } + 36x + 81 = 100 x^2\ ) .
Solution
\ ( \begin { array } { one hundred fifty }
4x^2 + 36x + 81 = 100 x^2 & \\
4x^2 + 36x + 81 − 100 x^2 = 0 & \qquad \text { 1. } \underline { \textbf { Zero } } \text {. Make one side zero } \\
{ \color { Cerulean } { 4x^2 } } +36x + { \color { Cerulean } { 81 } } − { \color { Red } { 100 x^2 } } & \qquad \quad \text { Observe three terms are perfect squares ( 2 positive, 1 negative ) } \\
( 2x+9 ) ( 2x+9 ) − 100 x^2=0 & \qquad \text { 2a. } \underline { \textbf { Factor } } \text {. Factor the perfect square binomial } \\
( 2x+9 ) ^2− ( 10x ) ^2=0 & \\
( 2x+9 − 10x ) ( 2x+9 + 10x ) =0 & \qquad \text { 2b. } \underline { \textbf { Factor } } \text {. Factor the difference of squares } \\
\begin { array } { three hundred } -8x+9=0 & \text { and } & 12x+9=0 \\
8x = 9 & & 12x = -9 \\
x = \dfrac { 9 } { 8 } & & x = -\dfrac { 9 } { 12 } \end { array }
& \begin { range } { fifty } \qquad \text { 3. } \underline { \textbf { property } } \text {. Set each component to 0 } \\
\quad \qquad \text { And solve } \\ \text { } \end { array } \\
\text { Solution Set : } \Big\ { \dfrac { 9 } { 8 }, -\dfrac { 3 } { 4 } \Big\ }
\end { align } \ )
Factoring using the sum or deviation of cubes formula to solve an equation will be discussed in the future section that includes the quadratic Formula and the Complete the Square technique .
Try It \ ( \PageIndex { 9 } \ )
Solve each equation by factoring .

 $$(3m−2)(2m+1)=0$$  $$3c^2=10c−8$$  $$25p^2=49$$ $$(2m+1)(m+3)=12m$$  $$123b=−6−60b^2$$  $$8x^3=24x^2−18x$$ $$64x^2 + 225 = 240x$$ $$3x^3 – 2x^2 – 12x + 8 = 0$$ $$9x^2 + 16 =27x + 64$$