Chapter 1.3, Problem 73E

To determine

To calculate: The factors of the trinomial ${\left(3x+2\right)}^2+8\left(3x+2\right)+12$ .

Answer to Problem 73E

The factors of the trinomial ${\left(3x+2\right)}^2+8\left(3x+2\right)+12$ are $3x+4$ and $3x+8$ .

Explanation of Solution

Given information:

The expression ${\left(3x+2\right)}^2+8\left(3x+2\right)+12$ .

Formula used:

The special product formula of perfect square to multiply a polynomial with itself which is mathematically expressed as,

  ${\left(A+B\right)}^2=A^2+2AB+B^2$

To find the factor of the trinomial of the form $a{x}^2+bx+c$ , find the numbers $p,q,r$ and $s$ such that sum of product of two numbers is equal to coefficient of $x$ $\left(ps+qr=b\right)$ , product of $p$ and $q$ is equal to the coefficient of $x^2$ $pq=a$ and product of $r$ and $s$ numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(px+r\right)$ and $\left(qx+s\right)$ are the factors of $a{x}^2+bx+c$ .

  $\begin{array}{c}a{x}^2+bx+c=pq{x}^2+\left(ps+qr\right)x+rs\\ =pq{x}^2+psx+qrx+rs\\ =px\left(qx+s\right)+r\left(qx+s\right)\\ =\left(px+r\right)\left(qx+s\right)\end{array}$

Calculation:

Consider the given expression ${\left(3x+2\right)}^2+8\left(3x+2\right)+12$ .

Recall the special product formula of perfect square to multiply a polynomial with itself which is mathematically expressed as,

  ${\left(A+B\right)}^2=A^2+2AB+B^2$

Apply it,

  $\begin{array}{c}{\left(3x+2\right)}^2+8\left(3x+2\right)+12={\left(3x\right)}^2+2\left(3x\right)\left(2\right)+2^2+8\left(3x\right)+8\cdot 2+12\\ =9x^2+12x+4+24x+16+12\\ =9x^2+36x+32\end{array}$

The simplified form of ${\left(3x+2\right)}^2+8\left(3x+2\right)+12$ is $9x^2+36x+32$ .

Recall that to find the factor of the trinomial of the form $a{x}^2+bx+c$ , find the numbers $p,q,r$ and $s$ such that sum of product of two numbers is equal to coefficient of $x$ $\left(ps+qr=b\right)$ , product of $p$ and $q$ is equal to the coefficient of $x^2$ $\left(pq=a\right)$ and product of $r$ and $s$ numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(px+r\right)$ and $\left(qx+s\right)$ are the factors of $a{x}^2+bx+c$ .

  $\begin{array}{c}a{x}^2+bx+c=pq{x}^2+\left(ps+qr\right)x+rs\\ =pq{x}^2+psx+qrx+rs\\ =px\left(qx+s\right)+r\left(qx+s\right)\\ =\left(px+r\right)\left(qx+s\right)\end{array}$

So, find $p,q,r$ and $s$ such that $pq=9$ , $ps+qr=36$ and $rs=32$ .

Here, $p=3$ , $q=3$ , $r=8$ and $s=4$ , so, $9x^2+36x+32$ will be factorized as,

  $\begin{array}{c}9x^2+36x+32=9x^2+\left(12+24\right)x+\left(8\right)\cdot \left(4\right)\\ =9x^2+12x+24x+32\\ =3x\left(3x+4\right)+8\left(3x+4\right)\\ =\left(3x+4\right)\left(3x+8\right)\end{array}$

Thus, the factors of trinomial ${\left(3x+2\right)}^2+8\left(3x+2\right)+12$ are $3x+4$ and $3x+8$ .