Is 9x^2 + 42x + 49 a perfect square trinomial and how do you factor it?

Mar 5, 2018

See below

Explanation:

We know that ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

In our case: 9x^2+42x+49=3^2·x^2+2·9·7·x+7^2=(3x+7)^2 This it's the same that

$\left(3 x + 7\right) \left(3 x + 7\right)$ is the factorization

Mar 5, 2018

If $9 {x}^{2} + 42 x + 49$ is a perfect square, then it must fit the pattern:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

To verify whether it fits the pattern, we set the first term of the pattern equal to the first term of the given trinomial:

${a}^{2} = 9 {x}^{2}$

Then we set the last term of the pattern equal to the last term of the given trinomial:

${b}^{2} = 49$

Taking the square root of both we obtain:

$a = 3 x$ and $b = 7$

We can use this information to check whether the middle term is true:

$2 a b = 42 x$

$2 \left(3 x\right) 7 = 42 x$

$42 x = 42 x$

The middle term is true, therefore, we conclude that the given trinomial is a perfect square.

We can use the left side of the pattern to factor the given trinomial:

${\left(3 x + 7\right)}^{2} = 9 {x}^{2} + 42 x + 49$