Question

Factorise 54x^2+42x^3-30x^4

Answer 1

54x^2+42x^3-30x^4
=6x^2(9+7x-5x^2)

Answer 2

Answer:

$6x^2(x - \frac{7 + \sqrt{229}}{10})(x - \frac{7 - \sqrt{229}}{10})$

Step-by-step explanation:

$54x^2 + 42x^3 - 30x^4 \\ \\ = 6x^2(9 + 7x - 5x^2) \\ \\ = 6x^2(-5x^2 + 7x + 9) \\ \\ Let's factorize \ -5x^2 + 7x + 9.$

$a = -5\ \ \ ; \ \ \ b = 7\ \ \ ; \ \ \ c = 9 \\ \\ \\ b^2 - 4ac \\ \\ = (7^2) - (4 \times -5 \times 9) \\ \\ = 49 - ( - 180) \\ \\ = 49 + 180 =\ 229$

$\\ \\ \\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \\ = \frac{- 7 \pm \sqrt{229}}{-10} \\ \\ = \frac{7 + \sqrt{229}}{10}\ \ \ OR \ \ \ \frac{7 - \sqrt{229}}{10}$

$\\ \\ \\ \therefore\ 6x^2(-5x^2 + 7x + 9) \\ \\ = 6x^2(x - \frac{7 + \sqrt{229}}{10})(x - \frac{7 - \sqrt{229}}{10}) \\ \\ \\ \therefore\ 54x^2 + 42x^3 - 30x^4 = \underline{\underline{6x^2(x - \frac{7 + \sqrt{229}}{10})(x - \frac{7 - \sqrt{229}}{10})}}$