How do you factor $y = x^{2} + 8 x - 9$ $y=x^2 + 8x - 9$ ?

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Answer 1 · 2016-04-02T13:54:01

$y = x^{2} + 8 x - 9 \to y = (x + 9) (x - 1)$ $y=x^2+8x-9" "->" "y=(x+9)(x-1)$

As there is no coefficient of $x^{2}$ $x^2$ so we are only looking for factors of 9 where their difference is 8.

Consider the factors 1 and 9

We know that $1 \times 9 = 9$ $1xx9=9$ . So that is ok!

We also know that $9 - 1 = 8$ $9-1=8$ So that is ok as well!

So we have the general form of $(x + ?) (x + ?)$ $(x+?)(x+?)$

The 8 from $8 x$ $8x$ is positive so for the difference we need 9 to be positive and the 1 to be negative so that we end up with $9 - 1 = 8$ $9-1=8$

$(x + 9) (x - 1)$ $color(blue)((x+9)color(brown)((x-1))$

Just as a check lets multiply out the brackets

$x (x - 1) + 9 (x - 1)$ $color(brown)(color(blue)(x)(x-1)color(blue)(+9)(x-1))$

$x^{2} - x + 9 x - 9$ $x^2-x+9x-9$

$x^{2} + 8 x - 9 Works ok!$ $x^2+8x-9color(red)(" Works ok!")$
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$y = x^{2} + 8 x - 9 \to y = (x + 9) (x - 1)$ $y=x^2+8x-9" "->" "y=(x+9)(x-1)$

How do you factor y=x2+8x−9y=x^2 + 8x - 9 ?