# Thread: Factoring With Respect to the Integers

1. ## Factoring With Respect to the Integers

I'm a little bit stuck on a few of my homework problems, and can't really find out how to do them in my text book. I don't really need help on both of them unless there're different ways to find the answer, I just need to learn how to solve them. Any help would be fantastic.
Mackenzie

"Completely factor the polynomial with respect to the integers"
1. x2-9x-22
2. 2x2-5x+2

2. Originally Posted by mackenzie25
I'm a little bit stuck on a few of my homework problems, and can't really find out how to do them in my text book. I don't really need help on both of them unless there're different ways to find the answer, I just need to learn how to solve them. Any help would be fantastic.
Mackenzie

"Completely factor the polynomial with respect to the integers"
1. x2-9x-22
2. 2x2-5x+2
given your expressions do you mean

1. $x^2-9x-22$
2. $2x^2-5x+2$

if so

1. $(x-11)(x+2)$
2. $(2x-1)(x-1)$ or $2(x-2)(x-.5)$

there is no roots since these are not equations just expressions

3. That's definitely what I was going for, but how did you factor it?
(Thanks! )

4. $ax^2 + bx + c$ where $a, b, c$, are real numbers

$
(mx+p)(nx+q)
$

where

$
mn = a, pq = c
$

and

$
pn + mq = b
$

5. $(5)(7)=(3+2)(3+4)=3(3+4)+2(3+4)=3^2+3(4)+2(3)+2(4)$

$=3^2+[3(4)+2(3)]+2(4)=35$

$Such\ an\ alternative\ allows\ us\ to\ expand\ (x+a)(x+b)$
$to\ x(x+b)+a(x+b)=x^2+(a+b)x+ab$

$Also\ (5)(7)=(9-4)(9-2)=9^2+9(-4-2)+(-4)(-2)=35$