#1

#2**+1 **

Thx so much do u know these by any chance ?

x^2+2x-8 x^2+x-8 x^2-4 ??? Pls ?

TandCforlife
Mar 27, 2017

#3**+1 **

Think of these as (x - a)(x - b) → x^{2} - (a+b)x +a*b

(1) Compare with x^{2} + 2x - 8

We must have a*b = -8 so what are the possible products:

-8 and 1

8 and -1

-4 and 2

4 and -2

We must also have -(a + b) = 2 so, using the above possible products:

-8 and 1 give -(-8 + 1) → 7 no!

8 and 1 give -(8 + 1) → -9 no!

-4 and 2 give -(-4 + 2) → 2 yes!

4 and -2 give -(4 - 2) → -2 no!

So we have x^{2} + 2x - 8 → (x - -4)(x - 2) → (x + 4)(x - 2)

.

Try this yourself with x^{2} - 4

x2 + x - 8 doesn't have nice integer roots. You can do this using the quadratic formula, for example.

.

Alan
Mar 27, 2017