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

2088 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

A curve has equation y = ax^2 + 3x, when x= -1, the gradient of the curve is -5. Work out the value of a.


Express (7+ √5)/(3+√5) in the form a + b √5, where a and b are integers.


Solving equations with unknown in both sides


Show that (n^2) + (n+1)^2 + (n+2)^2 = 3n^2 + 6n + 5, Hence show that the sum of 3 consecutive square numbers is always 2 away from a multiple of 3.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences