How do you factor $9 x^{2} + 9 x - 28$ $9x^2 + 9x -28$ ?

Question

2421 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-20T07:00:46

The answer is $= (3 x + 7) (3 x - 4)$ $=(3x+7)(3x-4)$

We use

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

We complete the squares

$9 x^{2} + 9 x - 28$ $9x^2+9x-28$

$= 9 (x^{2} + x) - 28$ $=9(x^2+x)-28$

$= 9 (x^{2} + x + \frac{1}{4}) - 28 - \frac{9}{4}$ $=9(x^2+x+1/4)-28-9/4$

$= 9 {(x + \frac{1}{2})}^{2} - \frac{121}{4}$ $=9(x+1/2)^2-121/4$

$= {(3 (x + \frac{1}{2}))}^{2} - {(\frac{11}{2})}^{2}$ $=(3(x+1/2))^2-(11/2)^2$

$= (3 (x + \frac{1}{2}) + \frac{11}{2}) (3 (x + \frac{1}{2}) - \frac{11}{2})$ $=(3(x+1/2)+11/2)(3(x+1/2)-11/2)$

$= (3 x + \frac{3}{2} + \frac{11}{2}) (3 x + \frac{3}{2} - \frac{11}{2})$ $=(3x+3/2+11/2)(3x+3/2-11/2)$

$= (3 x + \frac{14}{2}) (3 x - \frac{8}{2})$ $=(3x+14/2)(3x-8/2)$

$= (3 x + 7) (3 x - 4)$ $=(3x+7)(3x-4)$

How do you factor 9x2+9x−28 9x^2 + 9x -28?