For what value pf b will the polynomial p(x)=4x^3-3x^2+bx+6 have the same remainder when its divided by x-1 and by x-3?

Can someone help me get this started?

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- Sep 18th 2008, 03:36 PMjarr3dpolynomial
For what value pf b will the polynomial p(x)=4x^3-3x^2+bx+6 have the same remainder when its divided by x-1 and by x-3?

Can someone help me get this started? - Sep 18th 2008, 03:45 PMskeeter
do you know to perform synthetic division?

- Sep 18th 2008, 03:49 PMhwhelper
I think you can do synthetic division and equate the terms involving b. Or

p(x) = 4x^3-3x^2+bx+6

= 4x^2(x-1) + x^2 + bx + 6 (because -3x^2 = -4x^2 + x^2)

= 4x^2(x-1) + x(x-1) + (b+1)(x-1) + 6-b-1

Remainder when divided by x-1 is 5-b

Similarly

p(x) = 4x^2(x-3) + 9x(x-3) + (b+27)(x-3) - 3b-81+6

So remainder = 3b-75

If 5 -b = 3b - 75 then 80 = 4b and b = 20 (if I've got my arithmetic right). - Sep 18th 2008, 04:09 PMjarr3d
Yes i know synthetic division. and HWHelper. i did not understand your method

- Sep 18th 2008, 04:32 PMhwhelper
My method is simple. Sice the highest power in p(x) is x^3, consider

p(x) = (x-1)(ax^2+mx+c). It cannot be something like (x-1) * (zx^3+ax^2+mx+c) because there is no x^4 in p(x).

Expand it.

p(x) = ax^2(x-1) + mx(x-1) + c(x-1) = ax^3 - ax^2 + mx^2 - mx + cx - c

Equate coefficients. So

a = 4. (Coeff of x^3)

(-a + m) = -3 (Coeff of x^2) so -4 + m = -3 so m = 1.

Keep going.