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    

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Answered by Kishan T. Maths tutor

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