Chapter 1 on polynomial functions is finally over and starting from today we will look at polynomial equations and inequalities. In this lesson, we have learned how to reuse grade 4 mathematics – long division to divide a polynomial by a binomial and ‘remainder theorem’ to determine the remainder of the polynomial without dividing it.

So now let’s recall the long division we learned in grade 4:

Divide 6487 by 42.

Here, ‘42’ at the left will be the ‘**divisor**’;

‘6487’ at the centre will be the ‘

**dividend**’;

‘154’ at the top will be the ‘

**quotient**’ which the answer is needed to be in

**perfect columns**; and

‘19’ at the bottom and has a letter ‘

**R**’ representing it will the

**remainder**of the expression.

In that case, 6487=154(42)+19

**→That is, dividend=(divisor)(quotient)+remainder**

Let’s try some real ones:

1. Divide x³-6x²+4x+1 by x-3

[Important point to note: the terms of the equation must always be in descending order during calculation] (The subtract signs along the side of the long division are just to indicate that subtraction is going on between the expressions.)

1. Divide x³-6x²+4x+1 by x-3

[Important point to note: the terms of the equation must always be in descending order during calculation] (The subtract signs along the side of the long division are just to indicate that subtraction is going on between the expressions.)

Therefore, we can say:

**Method 1(provided by our teacher): x³-6x²+4x+1 = (x-3)(x²-3x-5)-14**

Or

Method 2 (provided our textbook): x³-6x²+4x+1 = (x²-3x-5)(-14/(x-3))

Or

Method 2 (provided our textbook): x³-6x²+4x+1 = (x²-3x-5)(-14/(x-3))

How about those with some terms missing?

2. Divide x⁴-2x²+1 by x+2 (we can notice that the ’x³’ term and the ‘x’ term are missing!)

**[In fact, we can insert terms with the value of ‘0’ into the equation. That is:**

x⁴-2x²+1 → x⁴+0x³-2x²+0x+1]

This time, since there is no remainder both methods 1 and 2 can get the same form of answer.

x⁴-2x²+1 → x⁴+0x³-2x²+0x+1]

Therefore, x⁴-2x²+1 divided by x+2 = (x+2)(x³-2x²+2x-4)

→ This is how long divisions of polynomials works.

Now, let’s try out the remainder theorem.

Before we start, let’s do an experiment.

a) Divide x³-4x²+3x-13 by x+4

b) Find f(-4)

b. f(x) = x³-4x²+3x-13

f(-4) = (-4)³-4(-4)²+3(-4)-13

=-153

→ Because both (a) and (b) get the same answer, we can deduce that:

**When a polynomial function P(x) is divided by (x-b), the remainder is P(b), and when it is divided by ax-b, the remainder is P(b/a), where a and b are integers and a≠0.**

Let's try one more example: Find the remainder of x³+2x²-13x+1 when it is divided by x-1

→ x-1=0

x=1

(1)³+2(1)²-13(1)+1

=1+2-13+1

=-9

Therefore, when x³+2x²-13x+1 is divided by x-1, the remainder will be -9.

This is the end of the lesson, hope you will find it helpful! :)

If there is anything with my entry, please feel free to tell me.

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