Simplify the following:
(2 x^2+3 x-20)/(x^2+6 x+8)
The factors of 8 that sum to 6 are 4 and 2. So, x^2+6 x+8 = (x+4) (x+2):
(2 x^2+3 x-20)/(x+4) (x+2)
Factor the quadratic 2 x^2+3 x-20. The coefficient of x^2 is 2 and the constant term is -20. The product of 2 and -20 is -40. The factors of -40 which sum to 3 are -5 and 8. So 2 x^2+3 x-20 = 2 x^2+8 x-5 x-20 = 4 (2 x-5)+x (2 x-5):
4 (2 x-5)+x (2 x-5)/((x+4) (x+2))
Factor 2 x-5 from 4 (2 x-5)+x (2 x-5):
(2 x-5) (x+4)/((x+4) (x+2))
((2 x-5) (x+4))/((x+4) (x+2)) = (x+4)/(x+4)×(2 x-5)/(x+2) = (2 x-5)/(x+2):
Answer: |(2x - 5) / (x + 2)
+14985 2x^2+3x-20/ x^2+6x+8 {nl} Factorisation
\(2x^2+3x-20 =0\)
a b c
\(x = {-3 \pm \sqrt{3^2-4*2*(-20)} \over 2*2}\)
\(x=-\frac{3}{4}\pm\sqrt{169}\)
\({\color{blue}x1=12.25 \ \ \ x2=-13.75}\)
\(x^2+6x+8\)
p q
\(x=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}\)
\(x=-3\pm\sqrt{9-8}\)
\( \( {\color{blue}\frac{2x^2+3x-20}{x^2+6x+8}}\) = \({\color{blue}\frac{(x-12.25)\times(x+13.75)}{(x+2)\times(x+4)}}\)\)
\( {\color{blue}\frac{2x^2+3x-20}{x^2+6x+8}}\) = \({\color{blue}\frac{(x-12.25)\times(x+13.75)}{(x+2)\times(x+4)}}\)
