Question #fef3e

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Question #fef3e

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Answer 1 · 2017-04-24T22:12:07

We can solve this by the factoring method.
[This video explains solving by factoring very well if you are still lost after this explanation.](https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring)

$x^{2} + 13 x + 42$ $x^2+13x+42$

Looking at this quadratic equation we need to find two numbers that when multiplied give us $42$ $"42"$ and when added give us $13$ $"13"$
So,

$x \cdot y = 42$ $x*y=42$

AND

$x + y = 13$ $x+y=13$

In this case, your two numbers are 6 and 7

$6 \cdot 7 = 42$ $6*7=42$

$6 + 7 = 13$ $6+7=13$

This means we have $(x + 6) (x + 7)$ $(x+6)(x+7)$
but our final answer is $x = - 6, - 7$ $x=-6, -7$

Think of it like this...
We have:

$x + 6 = 0$ $x+6=0$

AND

$x + 7 = 0$ $x+7=0$

So we need to move our numbers to the right side of the equals sign to get the value of x. That is why they are negative.

$x = - 6$ $x=-6$

$x = - 7$ $x=-7$

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