# How do you factor the expression 4x^2+5x-9?

Mar 19, 2016

Notice that the sum of the coefficients is zero and hence find factorisation:

$4 {x}^{2} + 5 x - 9 = \left(x - 1\right) \left(4 x + 9\right)$

#### Explanation:

Notice that the sum of the coefficients is $0$. That is: $4 + 5 - 9 = 0$.

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$4 {x}^{2} + 5 x - 9 = \left(x - 1\right) \left(4 x + 9\right)$

Mar 19, 2016

I llike the 'ac' method as it is a bulletproof method.
the product of coefficients 'a' and c is -36 so we need one +ve and one -ve.

now we systematically list the factors of 36 and we'll worry about the assignment of the +ve and -ve later.

1 x 36
2 x 18
3 x 12
4 x 9 and
6 x 6

SEARCH the listing for a 'pair' of factors with one +ve and one -ve so that we get a TOTAL of 'b' or total of 5 in this case.
This will occur with -4 and +9

so now we split up the middle term (the linear term) using these values:

$4 {x}^{2} - 4 x + 9 x - 9$
There are now four terms and each PAIR of terms will always have a common factor (this is why this method is 'bulletproof'.

$4 x \left(x - 1\right) + 9 \left(x - 1\right)$

that's pretty cool.
Now you have a sum of two terms and (x - 1) is common to both of them and can be factored again to:

(x - 1)( 4x + 9)
done
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