3x^3 -2x^2-147x+98=(ax-c)(bx+d)(bx-d). Find a, b, c, d if a, b, c, d are positive integers

(bx+d)(bx-d)=b^2x^2-d^2(ax-c)(bx+d)(bx-d)=(ax-c)(b^2x^2-d^2)=ab^2x^3-ad^2x-b^2cx^2+cd^2ab^2=3b^2c=2ad^2=147-cd^2=98From equations:a=3/b^2c=2/b^2d^2=49b^2Since a, b, c, d are positive integers, b must be 1. Then a=3, c=2, d=7

LK
Answered by Laura K. Further Mathematics tutor

7245 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

What is the equation of a circle with centre (3,4) and radius 4?


Given f(x)= 8 − x^2, solve f(3x) = -28


Point A lies on the curve y=3x^2+5x+2. The x-coordinate of A is 2. Find the equation of the tangent to the curve at the point A


Plot the graph of 1/x for x greater than 0.


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