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SectionFactoring Trinomials

Some trinomials of the form \(x^2+bx+c\) can be factored as a product of binomials. If a trinomial of this type factors, we have

\begin{equation*} \begin{aligned} x^2+bx+c \amp =(x+m)(x+n)\\ \amp =x^2+nx+mx+mn\\ \amp =x^2+(n+m)x+mn\text{.} \end{aligned} \end{equation*}

This gives us

\begin{equation*} b=n+m \text{ and }c=mn. \end{equation*}

In short, if the leading coefficient of a factorable trinomial is \(1\text{,}\) then the factors of the last term must add up to the coefficient of the middle term.

Example144

Factor \(x^2+12x+20\text{.}\)

Solution

Determine the factors of the last term whose sum equals the coefficient of the middle term. To do this, list all of the factorizations of \(20\) and search for factors whose sum equals \(12\text{.}\)

\(1\cdot 20\) \(1 +20=21\)
\(2\cdot 10\) \(\alert{2 +10=12}\)
\(4\cdot 5\) \(4 +5=9\)
Table145Factors of \(20\)

Choose \(20 = 2 \cdot 10\) because \(2 + 10 = 12\text{.}\) Therefore, \(12x=2x+10x\text{,}\) and we can write

\begin{equation*} x^2+\alert{12x}+20=x^2+\alert{2x+10x}+20. \end{equation*}

We factor the equivalent expression by grouping.

\begin{equation*} \begin{aligned} x^2+12x+20\amp = x^2+2x+10x+20\\ \amp = x(x+2)+10(x+2)\\ \amp = (x+10)(x+2) \end{aligned} \end{equation*}

Our factored form is \((x+10)(x+2)\text{.}\)

We check by multiplying the two binomials.

\begin{equation*} \begin{aligned} (x+2)(x+10)\amp =x^2+10x+2x+20\\ \amp = x^2+12x+20 \end{aligned} \end{equation*}

Note: We made a choice in the above problem to write \(12x=2x+10x\text{.}\) What would happen if we instead had written \(12x=10x+2x\text{?}\) Notice that in the following calculation the answer is equivalent.

\begin{equation*} x^2+\alert{12x}+20=x^2+\alert{10x+2x}+20. \end{equation*}

We factor the equivalent expression by grouping.

\begin{equation*} \begin{aligned} x^2+12x+20\amp = x^2+10x+2x+20\\ \amp = x(x+10)+2(x+10)\\ \amp = (x+2)(x+10)\\ \amp = (x+10)(x+2) \end{aligned} \end{equation*}

Since multiplication is commutative, the order of the factors does not matter.

\begin{equation*} \begin{aligned} x^2+12x+20 \amp = (x+2)(x+10)\\ \amp = (x+10)(x+2) \end{aligned} \end{equation*}

If the last term of the trinomial is positive, then either both of the constant factors must be negative or both must be positive.

Example146

Factor \(x^2y^2-7xy+12\text{.}\)

Solution

Determine the factors of the last term whose sum equals the coefficient of the middle term. To do this, list all of the factorizations of \(12\) and search for factors whose sum equals \(-7\text{.}\) Notice that in this case, both numbers must be negative.

\(-1\cdot -12\) \(-1 -12=-13\)
\(-2\cdot -6\) \(-2-6=-8\)
\(-3\cdot -4\) \(\alert{-3-4=-7}\)
Table147Factors of \(20\)

Choose \(12 = -3 \cdot -4\) because \(-3 -4 = -7\text{.}\) Therefore, \(-7xy=-3xy-4xy\text{,}\) and we can write

\begin{equation*} x^2y^2\alert{-7xy}+12=x^2y^2\alert{-3xy-4xy}+12. \end{equation*}

We factor the equivalent expression by grouping.

\begin{equation*} \begin{aligned} x^2y^2-7xy+12\amp = x^2y^2-3xy-4xy+12\\ \amp = xy(xy-3)-4(xy-3)\\ \amp = (xy-4)(xy-3) \end{aligned} \end{equation*}

Our factored form is \((xy-4)(xy-3)\text{.}\)

We check by multiplying the two binomials.

\begin{equation*} \begin{aligned} (xy-4)(xy-3)\amp =(xy)^2-3xy-4xy+12\\ \amp = x^2y^2-7xy+12 \end{aligned} \end{equation*}

The check is very important and is something you should plan to do on every single problem.

Keep in mind that some polynomials are what is called prime. For example, consider the trinomial \(x^2+3x+20\) and the factors of 20:

\begin{equation*} \begin{aligned} 20\amp =1\cdot 20\\ \amp = 2\cdot 10\\ \amp=4\cdot 5 \end{aligned} \end{equation*}

There are no integer factors of 20 whose sum is 3. Therefore, the original trinomial cannot be factored as a product of two binomials with integer coefficients. The trinomial is prime, that is, it cannot be factored with integer terms.

SubsectionFactoring Trinomials Using the \(ac\)-method

We will now extend this technique to factor trinomials whose leading coefficient is not 1. This technique, called the \(ac\)-method, again makes use of the grouping method for factoring four-term polynomials. If a trinomial in the form \(ax^2+bx+c\) can be factored, then the middle term, \(bx\text{,}\) can be replaced with two terms with coefficients whose sum is \(b\) and whose product is \(ac\text{.}\) This substitution results in an equivalent expression with four terms that can be factored by grouping.

Example148

Factor \(18x^2-31x+6\) using the \(ac\)-method.

Solution

Here \(a=18,\, b=-31\text{,}\) and \(c=6\text{.}\)

\begin{equation*} \begin{aligned} ac=18(6)=108 \end{aligned} \end{equation*}

Now we factor \(108\text{,}\) and search for factors whose sum is \(-31\text{.}\)

\begin{equation*} \begin{aligned} 108\amp = -1(-108)\\ \amp = -2(-54)\\ \amp= -3(-36)\\ \amp = \alert{-4(-27)}\\ \amp = -6(-18)\\ \amp = -9(-12) \end{aligned} \end{equation*}

In this case, the sum of the factors \(-27\) and \(-4\) equals the middle coefficient, \(-31\text{.}\) Therefore, \(-31x=-27x-4x\text{,}\) and we can write

\begin{equation*} \begin{aligned} 18x^2\alert{-31x}+6=18x^2\alert{-27x-4x}+6\text{.} \end{aligned} \end{equation*}

We factor the equivalent expression by grouping.

\begin{equation*} \begin{aligned} 18x^2-31x+6\amp = 18x^2-27x-4x+6\\ \amp = 9x(2x-3)-2(2x-3)\\ \amp = (2x-3)(9x-2) \end{aligned} \end{equation*}

Our factored form is \((2x-3)(9x-2)\text{.}\)

Example149

Factor \(4x^2y^2-7xy-15\) using the \(ac\)-method.

Solution

Here \(a=4,\, b=-7\text{,}\) and \(c=-15\text{.}\)

\begin{equation*} \begin{aligned} ac=4(-15)=-60 \end{aligned} \end{equation*}

Now we factor \(-60\text{,}\) and search for factors whose sum is \(-7\text{.}\)

\begin{equation*} \begin{aligned} -60\amp = 1(-60)\\ \amp = 2(-30)\\ \amp= 3(-20)\\ \amp = 4(-15)\\ \amp = \alert{5(-12)}\\ \amp = 6(-10) \end{aligned} \end{equation*}

In this case, the sum of the factors \(5\) and \(-12\) equals the middle coefficient, \(-7\text{.}\) Therefore, we replace \(-7xy\) with \(5xy-12xy\text{.}\)

\begin{equation*} \begin{aligned} 4x^2y^2-7xy-15\amp = 4x^2y^2+5xy-12xy-15\\ \amp =xy(4xy+5)-3(4xy+5)\\ \amp = (4xy+5)(xy-3) \end{aligned} \end{equation*}

Our factored form is \((4xy+5)(xy-3)\text{.}\)

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Example150

Factor \(5x^2+16xy+3y^2\text{.}\)

Solution

Here \(a=5,\, b=16y\text{,}\) and \(c=3y^2\text{.}\)

\begin{equation*} \begin{aligned} ac=5(3y^2)=15y^2 \end{aligned} \end{equation*}

Now we factor \(15y^2\text{,}\) and search for factors whose sum is \(16y\text{.}\)

\begin{equation*} \begin{aligned} 15y^2\amp = \alert{1y(15y)}\\ \amp = 1y^2(15)\\ \amp = 1(15y^2)\\ \amp= 3y(5y)\\ \amp = 3(5y^2)\\ \amp = 3y^2(5)\\ \end{aligned} \end{equation*}

In this case, the sum of the factors \(1y\) and \(15y\) equals the middle coefficient, \(16y\text{.}\) Therefore, we replace \(16xy\) with \(15xy+1xy\text{.}\)

\begin{equation*} \begin{aligned} 5x^2+16xy+3y^2\amp = 5x^2+15xy+1xy+3y^2\\ \amp =5x(x+3y)+1y(x+3y)\\ \amp = (5x+1y)(x+3y)\\ \amp = (5x+y)(x+3y) \end{aligned} \end{equation*}

Our factored form is \((5x+y)(x+3y)\text{.}\)

Example151

Factor \(18a^2b^2-ab-4\text{.}\)

Solution

Here \(a=18,\, b=-1\text{,}\) and \(c=-4\text{.}\)

\begin{equation*} \begin{aligned} ac=18(-4)=-72 \end{aligned} \end{equation*}

Now we factor \(-72\text{,}\) and search for factors whose sum is \(-1\text{.}\)

\begin{equation*} \begin{aligned} -72\amp = -1(72)\\ \amp = 1(-72)\\ \amp = -2(36)\\ \amp = 2(-36)\\ \amp = -4(18)\\ \amp = 4(-18)\\ \amp = -8(9)\\ \amp = \alert{8(-9)}\\ \amp = -24(3)\\ \amp = 24(-3)\\ \end{aligned} \end{equation*}

In this case, the sum of the factors \(8\) and \(-9\) equals the middle coefficient, \(-1\text{.}\) Therefore, we replace \(-ab\) with \(8ab-9ab\text{.}\)

\begin{equation*} \begin{aligned} 18a^2b^2-ab-4\amp = 18a^2b^2+(8ab-9ab)-4\\ \amp =18a^2b^2+(-9ab+8ab)-4\\ \amp =18a^2b^2-9ab+8ab-4\\ \amp =9ab(2ab-1)-4(2ab-1)\\ \amp = (9ab+4)(2ab-1)\\ \end{aligned} \end{equation*}

Our factored form is \((9ab+4)(2ab-1)\text{.}\)

Example152

Factor \(12y^3-26y^2-10y\text{.}\)

Solution

First, we will factor out the GCF.

\begin{equation*} \begin{aligned} 12y^3-26y^2-10y=2y(6y^2-13y-5) \end{aligned} \end{equation*}

After factoring out \(2y\text{,}\) the coefficients of the resulting trinomial are smaller and have fewer factors. We can factor the resulting trinomial using \(6=2(3)\) and \(5=(5)(1)\text{.}\) Notice that these factors can produce \(-13\) in two ways:

\begin{equation*} \begin{aligned} 2(-5)+3(-1)=-10-3\amp =-13\\ 2(\alert{1})+3(\alert{-5})=2-15\amp =-13 \end{aligned} \end{equation*}

Since the last term is \(-5\text{,}\) the correct combination requires the factors \(1\) and \(5\) to be opposite signs. Here we use \(2(1) = 2\) and \(3(-5) = -15\) because the sum is \(-13\) and the product of \((1)(-5) = -5\text{.}\)

\begin{equation*} \begin{aligned} 12y^3-26y^2-10y\amp =2y(6y^2-13y-5)\\ \amp = 2y(2y \: \: )(3y\: \: )\\ \amp = 2y(2y-5)(3y+1) \end{aligned} \end{equation*}

We check as follows.

\begin{equation*} \begin{aligned} 2y(2y-5)(3y+1) \amp =2y(6y^2+2y-15y-5)\\ \amp = 2y(6y^2-13y-5)\\ \amp =12y^3-26y^2-10y \end{aligned} \end{equation*}

The factor \(2y\) is part of the factored form of the original expression; be sure to include it in the answer.

Our factored form is \(2y(2y-5)(3y+1)\text{.}\)

Factor \(2x^2-9x-5\) using the \(ac\)-method.

Factor \(6x^2+7x-10\) using the \(ac\)-method.

Factor \(9x^2+6x-8\) using the \(ac\)-method.

If factors of \(ac\) cannot be found to add up to \(b\) then the trinomial is prime.