Factor completely...

a)3p^2-19pq-14q^2 b)16x^4-625y^4 x^2-y^2+8y-16

d)x^4-3x^3-13x^2+3x+12 e)4x^4+3x^3-4x^2-3x

Thanks!

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- Jan 1st 2008, 02:20 PMjohettFactoring polynomials...
Factor completely...

a)3p^2-19pq-14q^2 b)16x^4-625y^4 x^2-y^2+8y-16

d)x^4-3x^3-13x^2+3x+12 e)4x^4+3x^3-4x^2-3x

Thanks! - Jan 1st 2008, 03:07 PMtopsquark
Use the "ac method."

Multiply the leading coefficient by the constant term:

$\displaystyle 3 \cdot -14 = -42$

Now list all pairs of factors of -42:

1, -42

2, -21

3, -14

6, -7

7, -6

14, -3

21, -2

42, -1

Now which of these pairs sum to the coefficient of the linear term, in this case -19? The 2, -21 pair.

So rewrite -19 = 2 - 21 in your original expression:

$\displaystyle 3p^2-19pq-14q^2 = 3p^2 + (2 - 21)pq - 14q^2$

$\displaystyle = 3p^2 + 2pq - 21pq - 14q^2$

$\displaystyle = (3p^2 + 2pq) + (- 21pq - 14q^2)$

$\displaystyle = p(3p + 2q) - 7q(3p + 2q)$

$\displaystyle = (p - 7q)(3p + 2q)$

-Dan - Jan 1st 2008, 03:08 PMtopsquark
- Jan 1st 2008, 03:10 PMtopsquark
- Jan 1st 2008, 03:22 PMtopsquark
Recall that in the ac method we wrote the middle (linear in that case) coefficient as a sum, then we factored by grouping. I'm going to do the same thing here. Write -13 = -12 - 1:

$\displaystyle x^4-3x^3-13x^2+3x+12 = x^4 - 3x^3 + (-12 - 1)x^2 + 3x + 12$

$\displaystyle = x^4 - 3x^3 - 12x^2 - x^2 + 3x + 12$

$\displaystyle = (x^4 - 3x^3 - 12x^2) + (-x^2 + 3x + 12)$

$\displaystyle = x^2(x^2 - 3x - 12) - (x^2 -3x - 12)$

$\displaystyle = (x^2 - 1)(x^2 - 3x - 12)$

$\displaystyle = (x + 1)(x - 1)(x^2 - 3x - 12)$

where I have used the difference of squares formula again.

How do you see something like this? Well, for starters, you wouldn't have been given the problem if you couldn't factor it, so I looked for something like this. The rest is a function of experience. The more analytical way depends on a theorem which you likely have never heard of: the rational root theorem. This (more or less) says that if we have a polynomial $\displaystyle ax^n + bx^{n - 1} +~...~+ cx + d$ then the only possible linear factors with integer coefficients are of the form $\displaystyle ([\text{factor of a}] \cdot x - [\text{factor of d}] )$. This does not mean that the polynomial will have such factors, it only says that it might. You can use this to show that the above polynomial has factors of x + 1 and x - 1. However I suspect this technique is a little beyond your level since I doubt you know how to divide polynomials by long division.

-Dan - Jan 1st 2008, 03:24 PMtopsquark