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

Answered by Further Mathematics tutor

2385 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Solve the following simultanious equations: zy=28 and 2z-3y=13


Show that 2cos^2(x) = 2 - 2sin^2(x) and hence solve 2cos^2(x) + 3sin(x) = 3 for 0<x<180


Work out 7/(2x^2) + 4/3x as a single fraction in its simplest form.


Solve x^(-1/4) = 0.2


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning