# Factoring a polynomial

1. May 29, 2014

### tomtomtom1

hi all

I am stuyding how to factor equations such as :-

7x^2-9X-6

The problem i have is that it takes me too long to find 2 numbers whose sum is D and the same numbers whoes product is E.

Is there any way/tips/guide on how i can achieve this quickly?

2. May 29, 2014

### Ray Vickson

Have you studied how to solve quadratic equations? If so, you will know that the
quadratic equation $7 x^2 - 9 x - 6 = 0$ has roots $r_1$ and $r_2$, and that means that the expression $7 x^2 - 9 x - 6$ factors as $7(x - r_1)(x-r_2)$. You can use the standard quadratic root formulas to find $r_1$ and $r_2$.

BTW: the thing you are factoring is not an equation; it is an expression. An equation must have an '=' sign in it, and yours does not.

3. May 29, 2014

### tomtomtom1

what about:-

X^2+xy-2y^2
6x^2+5xy-6y^2

What is the process of factoring such expressions?

4. May 29, 2014

### Ray Vickson

Exactly the same as what I already explained.

5. May 29, 2014

### HallsofIvy

Staff Emeritus
$x^2+ xy- 2y^2= 0$
$x^2+ xy+ (1/4)y^2- (1/4)y^2- 2y^2= 0$
$(x+ (1/2)y)^2- (9/4)y^2= 0$
That's of the form "$a^2- b^2= (a- b)(a+ b)$with a= x+ (1/2)y and b= (3/2)y
so $x^2+ xy- 2y^2= (x+ (1/2)y+ (3/2)y)(x+ (1/2)y- (3/2)y)= (x+ 2y)(x- y)$

That's one way to do it- very "algorithmic". Another way, more "heuristic", is to note that the only way to factor $x^2$ (with integer coefficients) is (x)(x) while there are two ways to factor $-2y^2$ (again with integer coefficients) is (-2y)(y) or (2y)(-y). We try (-2y)(y): $(x- 2y)(x+ y)= x^2- 2xy+ xy- 2y^2= x^2- xy- 2y^2$. No, that's not what we want. So try (-y)(2y): $(x+ 2y)(x- y)= x^2+ 2xy- xy- 2y^2= x^2+ xy- 2y^2$. Yes, that's what we want.

$6x^2+ 5xy- 6y^2= 6(x^2+ (5/6)xy- y^2)= 6(x^2+ (5/6)xy+ (25/144)y^2- (25/144)y^2- y^2)= 6((x+ 5/12)^2- (169/144)y^2)= 6((x+ 5/12)^2- ((13/12)y)^2)= 6(x+ 5/12+ 13/12)(x+ 5/12- 13/12))= 6(x+ 18/12)(x- 8/12)= 6(x+ 3/2)(x- 2/3)= (2x+ 3)(3x- 2)$.

Or: 6=(3)(2) so we must have one of (3x- 3y)(2x+ 2y), (2x- 2y)(3x+ 3y), (3x- 2y)(2x+ 3y), or (2x- 3y)(3x+ 2y). Multiplying each of those out, we see that
$(3x- 3y)(2x+ 2y)= 6x^2+ 6xy- 6xy- 6y^2= 6x^2- 6y^2$
$(2x- 2y)(3x+ 3y)= 6x^2+ 6xy- 6xy- 6y^2= 6x^2- 6y^2$
$(3x- 2y)(2x+ 3y)= 6x^2+ 9xy- 4xy- 6y^2= 6x^2+ 5xy- 6y^2$
$(2x- 3y)(3x+ 2y)= 6x^2+ 4xy- 9xy- 6y^2= 6x^2- 5xy- 6y^3$
so that (3x- 2y)(2x+ 3y) is the correct factoring.

And we see that

Last edited: May 29, 2014
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