[PDF] L3 – 22 – Factor Theorem Lesson MHF4U - jensenmath




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[PDF] L3 – 22 – Factor Theorem Lesson MHF4U - jensenmath

In this section, you will learn how to determine the factors of a polynomial function of degree 3 or greater Part 1: Remainder Theorem Refresher

[PDF] L3 – 22 – Factor Theorem Lesson MHF4U - jensenmath 101352_622_ls_factor_theorem.pdf

L3 - 2.2 - Factor Theorem Lesson

MHF4U

Jensen

In this section, you will learn how to determine the factors of a polynomial function of degree 3 or

greater.

Part 1: Remainder Theorem Refresher

a) Use the remainder theorem to determine the remainder when í µ ( í µ ) =í µ ! +4í µ " +í µ-6 is divided by í µ+2

b) Verify your answer to part a) by completing the division using long division or synthetic division.

Factor Theorem:

í µ-í µ is a factor of a polynomial í µ(í µ) if and only if í µ ( í µ ) =0. Similarly, í µí µ-í µ is a factor of í µ(í µ) if and only if 𝑎 # $ 0=0.

Remainder Theorem: When a polynomial function

í µ(í µ) is divided by í µ-í µ, the remainder is í µ(í µ); and when it is divided by í µí µ-í µ, the remainder is í µ í±Ž ! " ) , where í µ and í µ are integers, and í µâ‰ 0.

Note: I chose synthetic since it is

a linear divisor of the form í µ-í µ. Example 1: Determine if í µ-3 and í µ+2 are factors of í µ ( í µ ) =í µ ! -í µ " -14í µ+24 í µ ( 3 ) =

Since the remainder is ___, í µ-3 divides evenly into í µ(í µ); that means í µ-3 _______________________ of í µ(í µ).

í µ ( -2 ) =

Since the remainder is not ____, í µ+2 does not divide evenly into í µ(í µ); that means í µ+2

____________________________ of í µ(í µ). Part 2: How to determine a factor of a Polynomial With Leading Coefficient 1 You could guess and check values of í µ that make í µ ( í µ ) =0 until you find one that works... Or you can use the Integral Zero Theorem to help.

Integral Zero Theorem

If í µ-í µ is a factor of a polynomial function í µ(í µ) with leading coefficient 1 and remaining coefficients that

are integers, then í µ is a factor of the constant term of í µ(í µ).

Note: Once one of the factors of a polynomial is found, division is used to determine the other factors.

Example 2: Factor í µ

! +2í µ " -5í µ-6 fully.

Let í µ

( í µ ) =í µ ! +2í µ " -5í µ-6

Find a value of í µ such that í µ

( í µ ) =0. Based on the factor theorem, if í µ ( í µ ) =0, then we know that í µ-í µ is a factor. We can then divide í µ(í µ) by that factor. The integral zero theorem tells us to test factors of _____

Test ________________________________. Once one factor is found, you can stop testing and use that factor to

divide í µ(í µ). í µ ( 1 ) = Since ________________, we know that ________________________ a factor of í µ(í µ). í µ ( 2 ) = Since ________________, we know that ____________________ a factor of í µ(í µ). You can now use either long division or synthetic division to find the other factors Method 1: Long division Method 2: Synthetic Division

Example 3: Factor í µ

% +3í µ ! -7í µ " -27í µ-18 completely.

Let í µ

( í µ ) =í µ % +3í µ ! -7í µ " -27í µ-18

Find a value of í µ such that í µ

( í µ ) =0. Based on the factor theorem, if í µ ( í µ ) =0, then we know that í µ-í µ is a factor. We can then divide í µ(í µ) by that factor. The integral zero theorem tells us to test factors of ________

Test ________________________________________________. Once one factor is found, you can stop testing and use that

factor to divide í µ(í µ).

Since ________________, this tell us that ____________ is a factor. Use division to determine the other factor.

We can now further divide í µ

! +2í µ " -9í µ-18 using division again or by factoring by grouping.

Method 1: Division

Method 2: Factoring by Grouping

Therefore,

í µ % +3í µ ! -7í µ " -27í µ-18=

Example 4: Try Factoring by Grouping Again

í µ % -6í µ ! +2í µ " -12í µ

Note: Factoring by grouping

does not always work...but when it does, it saves you time! Group the first 2 terms and the last 2 terms and separate with an addition sign.

Common factor within each group

Factor out the common binomial

Part 3: How to determine a factor of a Polynomial With Leading Coefficient NOT 1

The integral zero theorem can be extended to include polynomials with leading coefficients that are not 1.

This extension is known as the rational zero theorem.

Rational Zero Theorem:

Suppose í µ

( í µ ) is a polynomial function with integer coefficients and í µ= # $ is a zero of í µ(í µ) , where í µ and í µ are integers and í µâ‰ 0. Then, • í µ is a factor of the constant term of í µ(í µ) • í µ is a factor of the leading coefficient of í µ(í µ) • (í µí µ-í µ) is a factor of í µ(í µ)

Example 5: Factor í µ

( í µ ) =3í µ ! +2í µ " -7í µ+2

We must start by finding a value of

# $ where 𝑎 # $ 0=0. í µ must be a factor of the constant term. Possible values for í µ are: ____________ í µ must be a factor of the leading coefficient. Possible values of í µ are: ____________

Therefore, possible values for

# $ are: ________________________________

Test values of

# $ for í µ in í µ(í µ) to find a zero.

Since ________________________________________________ of í µ(í µ). Use division to find the other factors.

Example 6: Factor í µ

( í µ ) =2í µ ! +í µ " -7í µ-6

Part 4: Application Question

Example 7: When í µ

( í µ ) =2í µ ! -í µí µ " +í µí µ-2 is divided by í µ+1, the remainder is -12 and í µ-2 is a factor. Determine the values of í µ and í µ.

Hint: Use the information given to create 2 equations and then use substitution or elimination to solve.


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