How do you factor the expression $- 4 x^{2} - 9 x - 10$ $-4 x^2-9 x -10$ ?

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Answer 1 · 2018-07-25T18:05:48

$- 4 (x + \frac{9 - i \sqrt{79}}{8}) (x + \frac{9 + i \sqrt{79}}{8})$ $-4(x+\frac{9-i\sqrt79}{8})(x+\frac{9+i\sqrt79}{8})$

$- 4 x^{2} - 9 x - 10$ $-4x^2-9x-10$

$= - (4 x^{2} + 9 x + 10)$ $=-(4x^2+9x+10)$

In above quadratic polynomial

$B^{2} - 4 A C = {(9)}^{2} - 4 (4) (10) = - 79 < 0$ $B^2-4AC=(9)^2-4(4)(10)=-79<0$

The given polynomial has no real roots hence no real factors

Now, the complex roots of $4 x^{2} + 9 x + 10 = 0$ $4x^2+9x+10=0$ are given by quadratic formula as follow

$x = \frac{- 9 \pm \sqrt{9^{2} - 4 (4) (10)}}{2 (4)}$ $x=\frac{-9\pm\sqrt{9^2-4(4)(10)}}{2(4)}$

$= \frac{- 9 \pm i \sqrt{79}}{8}$ $=\frac{-9\pmi\sqrt{79}}{8}$

hence, the complex factors are given as

$- 4 x^{2} - 9 x - 10$ $-4x^2-9x-10$

$= - 4 (x - \frac{- 9 + i \sqrt{79}}{8}) (x - \frac{- 9 - i \sqrt{79}}{8})$ $=-4(x-\frac{-9+i\sqrt79}{8})(x-\frac{-9-i\sqrt79}{8})$

$= - 4 (x + \frac{9 - i \sqrt{79}}{8}) (x + \frac{9 + i \sqrt{79}}{8})$ $=-4(x+\frac{9-i\sqrt79}{8})(x+\frac{9+i\sqrt79}{8})$

How do you factor the expression −4x2−9x−10-4 x^2-9 x -10?