Thread: Can someone help me solve my Factorization problem.

1. Can someone help me solve my Factorization problem.

2*x ^2 - 13*x + 18 = 0

So far I get :

General Term : a*x^2 + b*x + c

Which would be factored into :

(m*x + p) * ( n*x + q)

Where

m*n = a
p*q = c and
P*n + m*q = b

Step 1 -

m*n = 2
p*q = 18
P*n + m*q = -13

Step 2 -

Possible integer multiples
m n (and obverse)
2 = 1 & 2

Possible integer multiples
p q (and obverse)
18 = 1 & 18
2 & 9
3 & 6

So I get

(2*x + p)*(x + q) now I think I get a - 13 and an 18 so lets try

(2*x - 3) * (x - 6)

I now get lost help anyone. ...

2. What 2 numbers when multiplied equal 36 and when added equal -13?.

$2x^{2}-9x-4x+18$

Group:

$(2x^{2}-9x)-(4x-18)$

Factor out something common:

$x(2x-9)-2(2x-9)=(x-2)(2x-9)$

Voila!

3. Thanks.

Thank you, very quick answer. Button pressed.

4. If $f(x)=ax^2+bx+c$ and $x_1,x_2$ are the real roots of $f$, then $f(x)=a(x-x_1)(x-x_2)$.
So, the roots of $2x^2-13x+18$ are $\displaystyle x_1=2,x_2=\frac{9}{2}$.
Then $\displaystyle 2x^2-13x+18=2(x-2)\left(x-\frac{9}{2}\right)=(x-2)(2x-9)$.