Find the factors of x^3−7x−6

To find the factors of x3-7x-6, a method called equating coefficients can be used.

Firstly, set the equation to zero and find a suitable number which satisfies the equation, in this case x=-1 satisfies the equation, therefore one factor is (x+1), therefore the cubic equation can be represented by:

(x+1)(Ax2+Bx+C)

Expanding this polynomial gives:

x(Ax2+Bx+C)+(Ax2+Bx+C)

Ax3+Bx2+Cx+Ax2+Bx+C

Ax3+(A+B)x2+(C+B)x+C

Equating the coefficients looks at the number before the variable and comparing it with the predicted equation.

Ax3=1x................................(1)

(A+B)x2=0x2...........................(2)

(C+B)x=-7x ............................(3)

C=-6 ......................................(4)

therefore A=1, C=-6, and B=-1

(x-1)(x2-x-6)

and this is a matter of factorising (x2-x-6) which is (x-3)(x-2)

and so the factorised form becomes:

(x-1)(x-3)(x-2)

Answered by Michael W. Maths tutor

11101 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Prove 2^n >n for all n belonging to the set of natural numbers


Calculate (7-i*sqrt(6))*(13+i*sqrt(6))


Differentiate this equation: xy^2 = sin(3x) + y/x


Solve the following simultaneous equations y + 4x + 1 = 0, y^2 + 5x^2 + 2x = 0


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