SOLUTION: I have been hitting and missing on most of the online tutorials and I need some interaction with a human for a better understanding of division of the following
2x^2 + 5x –
Algebra.Com
Question 27970: I have been hitting and missing on most of the online tutorials and I need some interaction with a human for a better understanding of division of the following
2x^2 + 5x – 12/ 9x^2 – 16 ÷ 2x^2 – 7x + 6/3x^2 – x – 4
in this case would I subtract the 2x^2 from the 9x^2 etc and simplify this before doing the division?
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
2x^2 + 5x – 12/ 9x^2 – 16 ÷ 2x^2 – 7x + 6/3x^2 – x – 4
= {[2x^2 + 5x – 12]/[9x^2 – 16]}X {[3x^2 – x – 4]/[2x^2 – 7x + 6]}
If you replace a division sign by an into sign then the fraction that comes after the division sign should be reciprocated and placed after the into sign.
={[2x^2 + (8x-3x) – 12]/[(3x+4)(3x-4)]}X {[3x^2+(-4x+3x)– 4]/[2x^2+(-4x-3x)+ 6]}
(While factorising a factorisable quadratic expression, the middle term is split as a sum of two parts in such a way that the multiplication of these two parts should give the product of the square term and the constant term)
={[(2x^2+ 8x)-3x– 12]/[(3x+4)(3x-4)]}X {[(3x^2-4x)+(3x– 4)]/[(2x^2-4x)-3x+ 6]}
={[2x(x+4)-3(x+4)]/[(3x+4)(3x-4)]}X {[x(3x-4)+1(3x– 4)]/[2x(x-2)-3(x- 2)]}
={[(x+4)(2x-3)]/[(3x+4)(3x-4)]}X {[(3x-4)(x+1)]/[(x-2)(2x-3)]}
=[(x+4)(2x-3)(3x-4)(x+1)]/[(3x+4)(3x-4)(x-2)(2x-3)]
(multiplying nr by nr and dr by dr
=[(x+4)(x+1)]/[(3x+4)(x-2)] [cancelling the common factors (2x-3)and(3x-4)]
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