(x+4)((x^2) - kx - 5) is expanded and simplified. The coefficient of the x^2 term twice the coefficient of the x term. Work out the value of k.

(x + 4) (x2 - kx - 5) = x3 - kx2 - 5x + 4x2 - 4kx - 20 = x3 + (4-k)x2 + (-5-4k)x - 20.The coefficients are: (4-k) of x2 and (-5-4k) of x. Now we can write the equation:(4-k) = 2(-5 - 4k) /expand4 - k = -10 - 8k /+8k4 + 7k = -10 /-47k = -14 /divide by 7k = -2

MF

Related Further Mathematics GCSE answers

All answers ▸

The coefficient of the x^3 term in the expansion of (3x + a)^4 is 216. Find the value of a.


Rationalise and simplify (root(3) - 7)/(root(3) + 1) . Give your answer in the form a + b*root(3) where a, b are integers.


Find the stationary point of 3x^2+7x


A curve is mapped by the equation y = 3x^3 + ax^2 + bx, where a is a constant. The value of dy/dx at x = 2 is double that of dy/dx at x = 1. A turning point occurs when x = -1. Find the values of a and b.