The Factor Theorem is closely related to the Remainder Theorem in solutions to polynomials. Basically it says that if ƒ(b/a) = 0, then (ax – b) is a factor of the polynomial ƒ(x). Conversely, if (ax – b) is a factor of ƒ(x), then ƒ(a/b) = 0 and ƒ(x) is divisible by (ax – b). Let me give an example.

Is (x – 1) a factor of 3x² – 2x – 1 ? Lets check by substituting x = 1 into the function. We get 3(1)² – 2(1) – 1 = 0. So (x – 1) is a factor. In fact the function can be factorized as (3x + 1)(x – 1) and you will also realise that (3x + 1) is a factor. As substituting x = −1/3 into the function will also give the result as zero.

Try these questions.

1) Find the value of k for which (x + 2) is a factor of 2x³ + kx² – 4x – 8.

2) If (x – k) is a factor of the expression kx³ + 9x² – 10kx -12, where k is a positive integer, find the value of k. Hence find the other factors of the expression.

Have fun and good luck.

Answers :

1) Substitute x = -2 into the expression. 2(-2)³ + k(-2)² – 4(-2) – 8 = 0. Resulting -16 + 4k + 8 – 8 = 0. Solving for k gives k = 4.

2) First substitute x = k into the expression ⇒ k(k)³ + 9(k)² – 10k(k) – 12 = 0

k↑4 – k² -12 = 0

(k² + 3)(k² – 4) = 0

k² = 4 or k² = -3 (this is not valid, so not a solution)

∴ k = 2 or -2 (this is dropped since K is given as positive integer). So only k = 2 is solution.

The expression now becomes 2x³ + 9x² -20x -12.

Next, using long division to divide 2x³ + 9x² – 20x -12 by (x – 2), resulting 2x² + 13x + 6. Then further factorise :

2x² + 13x + 6 = (2x + 1)(x+ 6) So the other factors are (2x + 1) and (x + 6)

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