How do you factor the trinomial $x^{2} - x - 42$ $x^2-x-42$ ?

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How do you factor the trinomial $x^{2} - x - 42$ $x^2-x-42$ ?

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Answer 1 · 2018-07-05T15:07:13

$(x - 7) (x + 6)$ $(x-7)(x+6)$

Explanation:

$the factors of - 42 which sum to - 1$ $"the factors of "-42" which sum to "-1$
$are - 7 and + 6$ $"are "-7" and "+6$

$x^{2} - x - 42 = (x - 7) (x + 6)$ $x^2-x-42=(x-7)(x+6)$

Answer 2 · 2018-07-05T15:29:21

$(x - 7) (x + 6)$ $(x-7)(x+6)$

Explanation:

Given: $x^{2} - x - 42$ $x^2-x-42$ .

With some $m a g i c$ $color(blue)(bb(magic))$ , notice how it equals:

$= x^{2} - 7 x + 6 x - 42$ $=x^2-7x+6x-42$

$= x (x - 7) + 6 (x - 7)$ $=x(x-7)+6(x-7)$

$= (x - 7) (x + 6)$ $=(x-7)(x+6)$

The $m a g i c$ $color(blue)(bb(magic))$ is that you pick two numbers, say $a$ $a$ and $b$ $b$ , whose sum equals the coefficient of the middle term and whose product equals the last terms.

In other words, we get:

$a b = - 42$ $ab=-42$

$a + b = - 1$ $a+b=-1$

Clearly, $- 7$ $-7$ and $6$ $6$ are the numbers.

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How do you factor the trinomial $x^{2} - x - 42$ $x^2-x-42$ ?

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Explanation:

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How do you factor the trinomial x2−x−42 x^2-x-42?

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How do you factor the trinomial $x^{2} - x - 42$ $x^2-x-42$ ?