Polynomials Part3

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Contents

Factor Theorem

\,x-a is a factor of P\left(x\right) if and only if P\left(a\right)=0

Factors Of Polynomial

Before we go to the proof, let us ask ourselves, what is a factor of polynomial? Or more simply, what is a factor?

Thus, for polynomial

What's the difference between the two? One doesn't have remainder, the other does.

If and only if

Whenever a theorem says if and only if, it means that the theorem works both ways. In this case, the theorem is actually a combination of

Note that a lot of theorem (but not all) actually works both ways, even though the if and only if is not written explicitly.

However, if we are asked to prove a theorem which states if and only if, we are are required to prove it both ways.

Proof

Usage

As with remainder theorem, we must learn how to use it properly.

To Prove Factor

Example : Show that \,x-1 is a factor of \,x^{2}-7x+6

Polynomial vs Equation

Example :

P\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x-3\right) This is a polynomial

Zeroes, Factors, Roots

Important Note

Example

P\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x-3\right)

Note that zeroes, factors and roots, though related, are different concepts. Giving one as the answer when asked for another will result in zero/less marks.

Relationship Between Zeroes, Factors, Roots

Writing Solutions

Take note that different type of questions will involve different ways of writing, even if we are dealing with the exact same numbers. Here we see some of the ways the questions can be asked for a relatively simple polynomial.

Exercise 8

1) Show that \,x-a is a factor of 2x^{2}-3\left(a-1\right)x+a\left(a-3\right)

2) Show that \,x+3 is factor of P\left(x\right)=x^{3}-x^{2}-17x-15. Hence, solve the equation P\left(x\right)=0

3) Show that -\frac{1}{2} is a zero of P\left(x\right)=2x^{3}-x^{2}-5x-2. Hence, factorize the polynomial completely.

4) Show that \,-2 is a root to the equation \,3x^{3}-7x^{2}-22x+8=0. Hence, find the other roots.

Interpreting Information into Equations

Example :

\,x-2 is a factor of P\left(x\right)

Thus, P\left(2\right)=0

Exercise 9

1) Complete the following table

Information Equation
\,x-3 is a factor of P\left(x\right) P\left(3\right)=0
P\left(x\right) have factors \,x-2 and \,2x+1 P\left(2\right)=0, P\left(-\frac{1}{2}\right)=0
P\left(x\right) is divisible by \left(x-a\right)\left(x-b\right) P\left(a\right)=0, P\left(b\right)=0
P\left(x\right) and G\left(x\right) have a common factor P\left(a\right)=G\left(a\right)=0(Assume common factor \,x-a)
P\left(x\right) can be divided by \,x-1 but leaves remainder \,2 when divided by \,3x+2 P\left(1\right)=0, P\left(-\frac{2}{3}\right)=2

2) The polynomial \,ax^{3}-5x^{2}+bx+7 has a factor \,2x+1 and leaves remainder \,25 when divided by \,x-2. Find the values of \,a and \,b.

3) P\left(x\right)=2x^{3}+ax^{2}+bx+30 is divisible by \,x+3 and \,x-2. Find the values of \,a and \,b.

4) P\left(x\right)=2x^{2}+3x+c and G\left(x\right)=x^{2}-2x+c (where c \neq 0). Find the value of \,c and that common factor.

Factorizing Polynomial

To fully factorize a cubic (or higher) polynomial, we need to obtain at least one factor first by trial-and-error method.

That is, we need to find a value \,a such that P\left(a\right)=0, \therefore x-a is a factor.

Trial and Error

Try factors of the constant

Examples

Note

1) Trial-and-error is useful to find one or two factors, so that the rest can be found by usual methods. We don't however, use trial-and-error to find ALL factors. This is because there are often factors that can't be found by this method

2) Quadratic polynomials that have complex roots generally does not need to be factored further (since if will involve complex factors)

Exercise 10

Factorize the following polynomial completely

1) \,2x^{3}-5x^{2}-4x+3

2) \,3x^{3}-16x^{2}+6x-5

3) \,x^{4}+3x^{3}-5x^{2}-3x+4

4) \,2x^{4}+x^{3}+7x-10

Long Questions

A few important things to keep in mind when doing multi-part questions

Hence/Thus

Whenever we see the word Hence or Thus in the question, it means

Hence, or Otherwise

If however, we see Hence, or otherwise, it means that we MAY use other methods. However, it is almost guaranteed that those other methods will be more time consuming, more susceptible to careless mistakes, and well, less fun than the hence method.

Example

Factorize P\left(x\right)=x^{3}-x completely. Hence, solve P\left(x\right)=6x-6

Exercise 11

1) P\left(x\right)=ax^{3}+5x^{2}+bx-28 has a factor \,x-2 and leaves remainder \,-10 when divided by \,x+3. Find the values of \,a and \,b. Factorize the polynomial completely.

2) P\left(x\right)=x^{4}-2x^{3}+ax^{2}+bx+36 is divisible by \,x-3 and \,x+2. Find the values of \,a and \,b. Factorize the polynomial completely.

3) Show that \,2x-1 is a factor of P\left(x\right)=2x^{3}-13x^{2}+26x-10. Hence, solve the equation P\left(x\right)=6x^{2}-23x+10

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