Show that the cubic function f(x) = x^3 - 7x - 6 has a root x = -1 and hence factorise it fully.

f(-1) = (-1)^3 - (-1)*7 - 6 = -1 + 7 - 6 = 0

Hence f(x) = (x+1)(x^2 + ax - 6)

Expand this out

f(x) = x^3 + ax^2 - 6x + x^2 + ax - 6 

      = x^3 + (a+1)x^2 + (a-6)x -6

By comparing co-efficients

a + 1 = 0

a - 6 = -7

a = -1

Thus 

f(x) = (x + 1)(x^2 - x - 6)

      = (x + 1)(x - 3)(x + 2)

JB
Answered by James B. Maths tutor

4102 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do you integrate sin^2(3x)cos^3(3x) dx?


A curve C has the equation y=5sin3x + 2cos3x, find the equation of the tangent to the curve at the point (0,2)


How do you find the possible values of cos(x) from 5cos^2(x) - cos(x) = sin^2(x)?


Integrate 3x^2 + 4/3 x^5 with respect to x


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences