# How do you factor x^2+14x=-49?

Apr 7, 2018

${\left(x + 7\right)}^{2} = 0$

#### Explanation:

${x}^{2} + 14 x = - 49 | + 49$
${x}^{2} + 14 x + 49 = 0$

$\text{Complete the perfect square}$

${\left(x + \frac{14}{2}\right)}^{2} + 49 - 49 = 0$
${\left(x + 7\right)}^{2} = 0$

Apr 7, 2018

${\left(x + 7\right)}^{2}$

Here's how I factored it:

#### Explanation:

${x}^{2} + 14 x = - 49$

To factor this, we first have to make one side equal to zero. We do this by adding $49$ to both sides of the equation:
${x}^{2} + 14 x + 49$

This is a perfect square trinomial: We know that:

• $14 = 2 \times 1 \times 7$
• $49 = 7 \times 7$

Therefore, the expression ${x}^{2} + 14 x + 49$ is equivalent to:
${\left(x + 7\right)}^{2}$

$- - - - - - - - - - - - - - - -$

To check our answer, let's see if ${\left(x + 7\right)}^{2}$ equals to the original expression:
$\left(x + 7\right) \left(x + 7\right)$
$x \cdot x = {x}^{2}$

$x \cdot 7 = 7 x$

$7 \cdot x = 7 x$

$7 \cdot 7 = 49$

And when we combine this we get:
${x}^{2} + 7 x + 7 x + 49$

We can still combine like terms ($7 x + 7 x$):
${x}^{2} + 14 x + 49$

Therefore, we know that ${\left(x + 7\right)}^{2}$ is correct.

Hope this helps!