Chapter 8.7, Problem 8CYU

To determine

To find: That the given polynomial is a perfect square trinomial, write yes or no. If yes then factor it.

Answer to Problem 8CYU

The trinomial $E=6x^2+30x+36$ is not a perfect square but can be factorized as $6x^2+30x+36=6\left(x+2\right)\left(x+3\right)$ .

Explanation of Solution

Given information:

The given trinomial is $6x^2+30x+36$ .

Concept used:

For polynomial $a{x}^2+bx+c$ is perfect square trinomial if it can be written as $a{x}^2+bx+c={\left(mx+n\right)}^2=m{x}^2+2\cdot mx.n+n^2$ , i.e. we can find two equal factors of $a\times c$ whose sum equals b .

Direct method for $a{x}^2+bx+c$ check whether $b^2=4ac$ , if yes then $a{x}^2+bx+c$ will be a perfect square. Also, if $b^2<4ac$ the trinomial cannot be factorized.

Calculations:

The given trinomial E (say) is $E=6x^2+30x+36$ .

We try to find two factors of $6\times 36$ such that their sum is 30 as follows:

  $6\times 36=2\times 3\times 2\times 2\times 3\times 3$

Since, from, factors $2\times 3\times 2\times 2\times 3\times 3$ we cannot find a pair of equal factors whose sum is 30 and product is $6\times 36$ so the trinomial $6x^2+30x+36$ is not a perfect square.

Moreover, $a=6$ , $b=30$ , and $c=36$ , then ${\left(30\right)}^2=900\ne 4\times 6\times 36=864$ so $6x^2+30x+36$ is not a perfect square.

Since, ${\left(30\right)}^2=900>864$ , the trinomial can be factorized.

  $6x^2+30x+36=6\left(x^2+5x+6\right)$

To factor the trinomial $\left(x^2+5x+6\right)$ , we have $6=2\times 3$ and $5=2+3$

  $6x^2+30x+36=6\left(x^2+2x+3x+6\right)$

  $\Rightarrow 6x^2+30x+36=6\left(x\left(x+2\right)x+3\left(x+3\right)\right)$

  $\Rightarrow 6x^2+30x+36=6\left(x+2\right)\left(x+3\right)$

Thus, the factored trinomial is $E=6x^2+30x+36=6\left(x+2\right)\left(x+3\right)$ .

Conclusion:

The trinomial $E=6x^2+30x+36$ is not a perfect square but can be factorized as $6x^2+30x+36=6\left(x+2\right)\left(x+3\right)$ .