In the expansion of (x-7)(3x**2+kx-3) the coefficient of x**2 is 0. i) Find the value of k ii) Find the coefficient of x. iii) write the fully expanded equation in terms of x

i) multiply out: 3x3+kx2-3x-21x2-7kx+21 simplify: 3x3+(k-21)x2+(-7k-3)x+21 the coefficient of x2 is 0 and therefore k-21=0 k=21.
ii)from i) the coefficient of x is (-7k-3) k=21 and therefore the required answer is (-7*21)-3 =-147-3 =-150 iii) from i) and ii): 3x3+(k-21)x2+(-7k-3)x+21 -> 3x**3-150x+21

Related Further Mathematics GCSE answers

All answers ▸

How can I find the equation of a straight line on a graph?


A particle is moving in a straight line from A to B with constant acceleration 4m/s^2. The velocity of the particle at A is 3m/s in the direction AB. The velocity of the particle at B is 18m/s in the same direction/ Find the distance from A to B.


The curve C is given by the equation x^4 + x^2y + y^2 = 13. Find the value of dy/dx at the point (-1,3). (A-level)


Work out the coordinates for the stationary point of y = x^2 + 3x + 4