In Algebra, there is an important theorem, The Remainder Theorem. This is used to solve polynomials, whereby a polynomial is divided by a linear divisor. Firstly, what is a polynomial ƒ(x) ? It is a function of a variable, such as x, in a descending order or ascending order. Examples are 4x³ + x² – 3x + 5 or 3 + 2x – x² + 6x³ , respectively. Examples of a linear divisor are (x + 2) , (3x – 4), etc.

If you divide, x³ + 2x² – 4x + 6 by x – 1 using the long division, you will get x² + 3x – 1 with a remainder of 5. I hope all you Sec. 3/4 students know how to do the long division. Now the Theorem says that ‘When a polynomial ƒ(x) is divided by a linear divisor ax-b (where a ≠ 0), the remainder is ƒ(b/a). So in the example stated, since divisor is x-1, the remainder will be ƒ(1). Basically substituting x = 1 into the polynomial will give (1)³ + 2(1)² – 4(1) + 6 = 5 (da-da !).

Try this with the following questions :

1) Find the remainder when 5x³ + 4x² -3x – 1 is divided by x-2.

2) Find the value of k if 2x³ – x² + (k + 1)x – 4k has a remainder of 38 when divided by x-3.

3) When 3x³ + kx² + hx + 6 is divided by x² – 3x + 2, the remainder is 5x + 4. Find the values of h and k.

I will provide answers in later posting. Meantime, have fun.

*Answers*

1) Substitute x = 2 into the polynomial and the result is 5(2)³ + 4(2)² -3(2) -1 = 49.

2) Substitute x = 3 into the poly. and solve for k.

2(3)³ – (3)² + (k + 1)(3) – 4k = 38 ⇒ 54 – 9 + 3k + 3 – 4k = 38 ⇒ k = 10.

3) Firstly the quadratic, x² – 3x + 2 = (x – 2)(x – 1). Then substituting x = 2 first and follow by x = 1 will provide 2 linear equations with 2 unknowns, k and h, which will then be easily solved, as follows.

3(2)³ + k(2)² + h(2) + 6 = 5(2) + 4 ⇒ 24 + 4k +2h + 6 = 14 ⇒ 4k + 2h = – 16 or 2k + h = -8.

3(1)³ + k(1)² + h(1) + 6 = 5(1) + 4 ⇒ 3 + k + h + 6 = 9 ⇒ k + h = 0 or k = -h.

substitute k = -h into 2k + h = -8 ⇒ -2h + h = -8 ⇒ -h = -8 or h = 8 and k = -8

So, not too difficult, I hope. Enjoy your long weekend, Labour Day holidays.

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