Polynomials Part2

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Remainder Theorem

When P\left(x\right) is divided by \,x-a, its remainder is P\left(a\right)

Proof

When P\left(x\right) is divided by \,x-a,

Divided by ax+b

Similarly, we will get

Usage

So what's so great about the Remainder Theorem? All it say's is that the remainder is P\left(a\right) when P\left(x\right) is divided by \,x-a.

Well, note that

Important Notes

As stated earlier, this subtopic is where HOW you write is as important, or even more important than WHAT you write. Also, make sure you understand WHY it needs to be written that way. For now, keep these two things in mind

Examples

Common Mistakes

Consider the following sample solution

Now consider the following (even more common)

Now try doing the next example

Exercise 4

Find the remainder when P\left(x\right)=x^{2}-5x+6 is divided by the following divisor

Divisor Remainder
\,x-1 P\left(1\right) \,2
\,x+2 P\left(-2\right) \,20
\,x-3 P\left(3\right) \,0
\,2x-1 P\left(\frac{1}{2}\right) \frac{15}{4}
\,3x+2 P\left(-\frac{2}{3}\right) \frac{88}{9}
\,4x-3 P\left(\frac{3}{4}\right) \frac{45}{16}
\,5x+1 P\left(-\frac{1}{5}\right) \frac{176}{25}

Interpreting Information into Equations

Example :

P\left(x\right) gives remainder \,5 when divided by \,x-2

Thus, P\left(2\right)=5

Exercise 5

Complete the following table

Information Equation
P\left(x\right) leaves remainder \,1 when divided by \,x-3 P\left(3\right)=1
When P\left(x\right) is divided by \,x+a its remainder is \,b P\left(-a\right)=b
P\left(x\right) gives remainder \,-20 and \,5 when divided by \,x+2 and \,2x-3 respectively. P\left(-2\right)=-20,P\left(\frac{3}{2}\right)=5
P\left(x\right) is divisible by \,x-4 P\left(4\right)=0
P\left(x\right) have the same remainder when divided by \,2x-3 and \,3x+1 respectively P\left(\frac{3}{2}\right)=P\left(-\frac{1}{3}\right)
P\left(x\right) and G\left(x\right) have the same remainder when divided by \,x-3 P\left(3\right)=G\left(3\right)

Solving Questions

The typical simple question involves

Simultaneous Equations

When solving simultaneous equation

Getting rid of fractions



Example

\frac{1}{16}a+\frac{1}{2}b=-6 & \,3a-2b=50

More on Checking Solutions

After solving questions, we check that our answers satisfies the original given condition/equation in the question, and not some equations in the middle part.

Example

\,ax^{3}-7x^{2}+bx+7 leaves remainder \,\frac{1}{3} and \,11 when divided by \,3x+2 and \,x-2 respectively. Find the values of \,a and \,b respectively.

Sample solution:

Analysis:

General Guidelines:

Exercise 6

1) \,x^{3}+ax^{2}-5x+b leaves remainder \,1 and \,7 when divided by \,x-2 and \,x+1 respectively. Find the values of \,a & \,b.

2) \,ax^{3}-5x^{2}+bx+7 leaves remainder \,12 and \,-29 when divided by \,2x-1 and \,2x+3 respectively. Find the values of \,a & \,b.

3) \,3x^{3}-2x^{2}+ax+2 leaves a remainder \,b and \,-4b when divided by \,3x+1 and \,x+2 respectively. Find the values of \,a & \,b.

4) \,x^{2}+ax-5 and \,ax^{2}-5x-a leaves the same remainder when divided by \,x+3. Find value of \,a

5) \,5x^{2}-2x+17 leaves the same remainder when divided by \,x+a and \,x-3a respectively. Find \,a, where a \neq 0

Quadratic Divisor

When P\left(x\right) is divided by a quadratic divisor, its remainder is a linear function.

Thus, to find its remainder without long division, we can't use remainder theorem directly. We will however, use the same steps as those in the proof of the theorem.

Note : The most important step is writing the above equation.

Examples

Exercise 7

1) Without using long division, find the remainder of \,2x^{3}+x^{2}-16x-4 when divided by

2) \,ax^{3}+4x^{2}+bx-7 gives remainder \,13x-5 when divided by \left(x-1\right)\left(x+2\right). Find values of \,a and \,b.

3) f\left(x\right) leaves remainder \,-21 and \,1 when divided by \left(x+5\right) and \left(2x-1\right) respectively. Find remainder when f\left(x\right) is divided by \,2x^{2}+9x-5

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