anonymous 4 years ago how do i simplify 4x^2-7x+3 divided by x^2+5x-6?

1. ash2326

we have $\frac{4x^2-7x+3}{x^2-5x+6}$ let's divide (x^2-5x+6)*4=4x^2-20x+24 rewriting the numerator as 4x^2-7x+6=(4x^2-20x+24)+13x-18 so we have now $\frac{(4x^2-20x+24)+13x-18 }{x^2-5x+6}$ if we simplify we get $4+\frac{13x-18 }{x^2-5x+6}$

2. anonymous

is there more simplifying? because thats not one of my options

3. anonymous

i wrote the question wrong, the denomenator is supposed to be x^2+5x-6

4. ash2326

then 4(x^2+5x-6)=4x^2+20x-24 so 4x^2-7x+3=(4x^2+20x-24)-27x+27 so we get $4+\frac{-27x+27}{x^2+5x-6}$

5. anonymous

these are my options

6. anonymous

$x^2 + 5x - 6 = (x-1)(x+6)$ $4x^2-7x+3 = 4x^2-4x-3x+3=4x(x-1)-3(x-1)=(4x-3)(x-1)$ So cancelling (x-1) you can see the answer

7. anonymous

$\frac{4x-3}{x+6}$