Write down three linear factors of f(x) such that the curve of f(x) crosses the x axis at x=0.5,3,4. Hence find the equation of the curve in the form y = 2(x^3) + a(x^2) + bx + c.

the curve crosses the graph at the x axis at 0.5,3 and -4 so f(0.5)=0, f(3)=0,f(-4)=0. All linear factors are the form of g(x)=(x-a) so 0=g(0.5)=0.5-a. rearranging we get that a=0.5 ie g(x)=x-0.5 similarly the other linear factors are (x-3),(x+4) so f(x) is the product of three linear factors and so f(x)=(x-0.5)(x-3)(x+4) expanding f(x)=(x-0.5)[(x-3)(x+4)] =(x-0.5)[x(x+4)-3(x+4)] =(x-0.5)[x2+4x-3x-12] =(x-0.5)(x2+x-12)  =x(x2+x-12)-0.5(x2+x-12)  =x3+x2-12x-0.5x2-0.5x+6 =x3+0.5x2-12.5x+6   now we can multiply the coeffiecient by 2 to get the desired form f(x)=2x3+x2-25x+12    

KT
Answered by Kishan T. Maths tutor

6609 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The mass of a substance is increasing exponentially. Initially its mass is 37.5g, 5 months later its mass is 52g. What is its mass 9 months after the initial value to 2 d.p?


Express 3sin(2x) + 5cos(2x) in the form Rsin(2x+a), R>0 0<a<pi/2


Differentiate 3x^(2)+xy+y^(2)=12 with respect to x


How would you expand (x+5y)^5?


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:

© 2026 by IXL Learning