(x + a)(x + 3)(2x+1) = bx^3 + cx^2 + dx -12, find the values of a, b, c and d.

This question is an example of expanding algeba and equating coefficients. If the brackets on the left-hand side are expanded the unknown values can be found from the coefficients on the right-hand side.Step 1) Expand the brackets(x2 + (3 + a)x + 3a)(2x + 1) = bx 3 + cx2 + dx - 122x3 + (6 + 2a)x2 + 6ax + x2 + (3 + a)x + 3a = bx 3 + cx2 + dx - 12Now expanded, the equation should be simplified.2x3 + (7 + 2a)x2 + (7a + 3)x + 3a = bx 3 + cx2 + dx - 12Now one by one, the variables can be calculated.a = -4, Hence:2x3 - x2 - 25x - 12b = 2, c = -1, d = -25(This is an example question from an OCR GCSE higher paper June 2018)

Answered by Joe S. Maths tutor

4992 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

A solution to the equation 2x^2-3x-17=0 lies between 2&3 use method of trail and improvement to find the solution


Simplify the expression: (2a + a)x(5a - a)


n is an integer such that 3n + 2 < 14 and 6n/(n^2+5) > 1. Find all possible values of n.


How do you find the area of a sector of a circle if you know the radius and the angle in the centre?


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 2013–2024

Terms & Conditions|Privacy Policy