Question

Factor this polynomial expression: 2(2x-1)^2-9(2x-1)+4

Answer 1

let z = 2x-1

which means you'll go from 2(2x-1)^2-9(2x-1)+4 to 2z^2-9z+4

Answer 2

are you able to factor 2z^2-9z+4 ?

Answer 3

I did that, and I'm normally excellent at this, but I'm having a mental block and can't factor that even with the z's.

Answer 4

use the AC method

A = 2
B = -9
C = 4

A*C = 2*4 = 8

find two numbers that multiply to 8 and also add to B = -9

Answer 5

hint:

here are all the ways to multiply two whole numbers (positive or negative) to get 8

1*8 = 8
2*4 = 8
(-1)*(-8) = 8
(-2)*(-4) = 8

Answer 6

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Answer 7

@jim_thompson5910 I tried it and I got (4x-3)(2x-9) when the answer is supposed to be (4x-3)(2x-5)...

I did 2z^2-9z+4, then I did the AC method to get (2z-1)(z-8). Then I replaced the z with the actual equation and got [2(2x-1)-1][2x-1-8]. I then simplified to (4x-3)(2x-9)...

Answer 8

(2z-1)(z-8) isn't the correct factorization for 2z^2-9z+4

Answer 9

2x4 is 8, so what is the correct factorization?

Answer 10

2z^2-9z+4

2z^2-1z-8z+4 ... note: -1 and -8 add to -9 and multiply to +8

(2z^2-1z)+(-8z+4)

z(2z-1)-4(2z-1)

(z-4)(2z-1)

Answer 11

Oh it's -2 and -4. But how am I supposed to know it's that and not -1 and -8 by just eyeballing it?

Answer 12

(2z-1)(z-8) expands out to...


(2z-1)(z-8)

2z*(z-8) - 1*(z-8)

2z^2 - 16z - z + 8

2z^2 - 17z + 8

-------------------------------------------------------

while (z-4)(2z-1) expands out to ...


(z-4)(2z-1)

z*(2z-1) - 4*(2z-1)

2z^2 - z - 8z + 4

2z^2 - 9z + 4

Answer 13

Sorry I lost connection. Okay, I understand. Is there a way of eyeballing it to know it's not -1 and -8 without the calculation?

Answer 14

Here's another way to factor 2z^2-9z+4

use the quadratic formula to solve 2z^2-9z+4 = 0 to get these two solutions: z = 4 or z = 1/2

use these two solutions to factor

z = 4 or z = 1/2

z = 4 or 2z = 1

z-4 = 0 or 2z-1 = 0

(z-4)*(2z-1) = 0


notice how I'm working the zero product property in reverse

Answer 15

Alright, thanks, but like I said is there a way of eyeballing what I said?

Answer 16

I'm not sure what you mean when you say "Is there a way of eyeballing it to know it's not -1 and -8 without the calculation?"

the two numbers are -1 and -8. I don't know what you mean when you have "not" in there.

Answer 17

The factor is -2 and -4 not -1 and -8, so I was wondering if there was a way of telling that it's the first and not the latter without expanding them both like you did a few comments ago?

Answer 18

Factoring 2z^2-9z+4 using the AC method

A = 2
B = -9
C = 4

A*C = 2*4 = 8

find two numbers that multiply to 8 and also add to B = -9

those two numbers are -1 and -8 (in either order). You arrive at this pair of numbers through guess and check. A list like you see below may help

1*8 = 8
2*4 = 8
(-1)*(-8) = 8
(-2)*(-4) = 8

-------------------------------------------------------

So again, the pair of numbers is -1 and -8. Use them to break up -9z and then factor by grouping


2z^2-9z+4

2z^2-1z-8z+4

(2z^2-1z)+(-8z+4)

z(2z-1)-4(2z-1)

(z-4)(2z-1)

Answer 19

I'm still not sure why you're saying "The factor is -2 and -4 not -1 and -8"

Answer 20

Somehow I skipped a step and it seemed way easier. Now it's more difficult but is correct! Thank you!