Finding Unknown Coefficients

Video lesson · from 22:252 worked examples

One lesson video covers all of Polynomials & Quadratics, so it opens at 22:25 for this topic — not from the beginning.

Consider a polynomial with some unknown coefficients, such as $x^3+2px^2-px+4$ , where $p$ is a constant.

If we divide the polynomial by $x-h$ , then we will obtain an expression for the remainder in terms of the unknown constants.

If we already know the value of the remainder then we can solve for the unknown constants.

A factor means remainder zero: that is the equation you solve for the unknown. Set the last entry of the synthetic division to 0.
Two unknowns need two conditions: the question will give you both — usually two factors, or a factor and a remainder. Form both equations and solve simultaneously.
Do not stop at the unknown: most questions then ask you to factorise fully or find the roots. Read to the end.
Substitute your value back: a quick check that the remainder really is zero catches arithmetic slips.

Example 1

Given that $x-3$ is a factor of $f(x)=x^3-x^2+px+24$ , find the value of $p$ .

Since $x-3$ is a factor, by the factor theorem, $f(3) = 0$ .

f(3) = (3)^3 - (3)^2 + p(3) + 24 = 0

27 - 9 + 3p + 24 = 0

42 + 3p = 0

3p = -42

p = -14

Example 2

When $f(x)=px^3+qx^2-17x+4q$ is divisible by $x-2$ , the remainder is 6.

$(x-1)$ is a factor of $f(x)$ .

Find the values of $p$ and $q$ .

From the first statement, $f(2) = 6$ :

p(2)^3 + q(2)^2 - 17(2) + 4q = 6

8p + 4q - 34 + 4q = 6

8p + 8q = 40

Dividing by 8: $p + q = 5 \quad \textbf{(1)}$

From the second statement, $f(1) = 0$ :

p(1)^3 + q(1)^2 - 17(1) + 4q = 0

p + q - 17 + 4q = 0

p + 5q = 17 \quad \textbf{(2)}

Subtracting equation (1) from equation (2):

(p + 5q) - (p + q) = 17 - 5

4q = 12 \implies q = 3

Substitute $q = 3$ into equation (1):

p + 3 = 5 \implies p = 2

So $p = 2$ and $q = 3$ .