Question

Factor Completely:

18x^2-35x+12

I know how to factor, but I am not sure of the rule if products of 12 don't equal 35.

Answer 1

x(18x-35)+12

(2x-3)(9x-4)

Answer 2

So you factored out the x, left the 12 by itself, but then how did you get that?

Answer 3

Ah, but that
\[18x^{2} \]
changes everything.

Answer 4

Okay, I see what you did there. I think, anyway. You pulled out x(18x-35) + 12. Well, actually, I still don't get how you got (2x-3)(9x-4)

Answer 5

Since it's not a monic quadratic, you're trying to find two numbers that multiply to be 216 rather than 35, and add to be -35. [[216 comes from 18*12]]

-27 and -8 work, so split 35x into -27x and -8x

18x² - 35x + 12
= 18x² - 27x - 8x + 12
= 9x(2x - 3) - 4(2x - 3)
= (9x - 4) (2x-3)

Answer 6

So you can multiple the coefficient of a by c, then find something in there that equals 35?

Answer 7

*multiply

Answer 8

Just pick some representation of 18 as the product of two factors:

18 & 1
9 & 2
6 & 3

Since the middle term is negative and the first and last terms are positive you know you have something in the form

(ax - c)(bx - d)

And you know that a and b must be 2 numbers on that list.

So you have
c*d = 12
ad + bc = 35

And you have 3 different sets you can test for a/b to solve for the c and d using the system above.

Answer 9

To factorize the polynomial ax² + bx + c, you need to find 2 numbers (let them be m and n) that satisfy these two conditions:
1) m*n = a*c
2) m + n = b

Then you split bx into mx and nx and factorize by grouping in pairs

Answer 10

Okay, I get that, ty both vm :D

Answer 11

no problem =)