Factorising quadratics.

Question

If i am factorising x²-x-72=0, I know i need to put it in the format of (x-n)(x+m) and i know 9x8 is 72 but i dont know which way to but the numbers in the brackets

Answer 1

Hello,

Yes, you are correct in saying that 9*8=72. In order to determine how to put the numbers in the brackets, you need to determine in which order the numbers must be added together to get -x. Since the b term is negative, the 9 must be negative. Since you want to end up with -x, you need to add 8 to the -9 to obtain -1. Therefore, x^2-x+72=0, you would factor it as (x-9)(x+8).

Answer 2

x^2-x-72=0

 

I know i need to put it in the format of

 

(x-n)(x+m)

 

= x^2 + (m-n)x – nm

 

Compare to x^2-x-72 and:

 

m-n = -1, nm = 72 = 8(9)

 

9-8 = +1 X

8-9 = –1 √

 

So x^2-x-72

=(x-9)(x+8)

 

check:

 

(x-9)(x+8) =

 

x^2 + 8x – 9x – 72 =

 

x^2 – x – 72 √

Answer 3

1(x)    -9

 

1(x)      8

 

You can write like the above, since 1=1*1, -72=-9*8, make sure that the sum of the products of the numbers in the diagonal line is equal to the coefficient of the term of first degree (that is the coefficient of -x). For this factorization, 1*8+1*(-9)=-1. Then the factorization is (x-9)(x+8), that is the product of the first line and the second line.

 

If you have question, please feel free to ask.

 

YJ