Given a polynomial f(x) where f(x) = x^3 + ax^2 + 3x + b and ( x− 1) is a factor of f(x) and f ′(x)= 12 at point ( 2,4) find the values of a and b hence factorise f(x) completely


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Question Number 38910 by Rio Mike last updated on 01/Jul/18

$G i v e n a p o l y n o m i a l f (x) w h e r e$ $f (x) = x^{3} + a x^{2} + 3 x + b$ $a n d (x - 1) i s a f a c t o r o f f (x)$ $a n d f^{'} (x) = 12 a t p o i n t (2, 4)$ $f i n d t h e v a l u e s o f a a n d b$ $h e n c e f a c t o r i s e f (x) c o m p l e t e l y$
	Answered by ajfour last updated on 01/Jul/18

	$1 + a + 3 + b = 0$ $f^{'} (x) = 3 x^{2} + 2 a x + 3$ $f^{'} (2) = 12 + 4 a + 3 = 12$ $\Rightarrow a = - \frac{3}{4}; b = - \frac{13}{4}$ $f (x) = (x - 1) (x^{2} + p x + q)$ $\Rightarrow - q = b = - \frac{13}{4}$ $c o e f f i c i e n t o f x^{2}$ $p - 1 = a = - \frac{3}{4}$ $f (x) = (x - 1) (x^{2} + \frac{x}{4} + \frac{13}{4}) .$
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