Could you factorize the following? 2(2x-1)^2 - 3(x-3)^2 = 5(x+3)(2x-1)I'm sorry. That was meant to be 3(x+3)^2

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neela | High School Teacher | (Level 3) Valedictorian

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2(2x-1)^2 - 3(x-3)^2 = 5(x+3)(2x-1)

We solve the above equation using factors :

2(2x-1)^2-3(x-3)^2 = 5(x+3)(2x-1)

2(4x^2-4x+1) -3(x^2-6x+9) = 5(2x^2-x+6x-3)

(8-3)x^2+(-8+18)x +2-27 = 10x^2+25x-15

5x^2+10x-25 = 10x^2+25x-15

0 = (10-5)x^2+(-10+25)x+25-15

5x^2+15x+10 = 0

Divide by 5:

x^2+3x+3

(x+1)(x+2) = 0

x+1 = 0. Or (x+2 = 0.

x=-1. Or x = -2..

If x-3 is to be replaced by x+3 as suggested in he next page posted:

Then instead of -3(x-6x+9) , we must change to  -3(x^2+6x+9).

The correction is  -36x^2 instead of 26x^2.

This leads to 10x^2+51x +10 = 0

10x^2 +50x+x+10 = 0

x(10x+1)+1(10x+1) = 0

(10x+1)(x+1) = 0

10x+1 = 0. Or x+1 = 0

10x = -1 or x = -1/10.

x+1 = 0 gives x = -1

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