factoring polynomials

FACTORING FACTORING POLYNOMIALSPOLYNOMIALS

Main ideas:Main ideas:

What is the numerical value of a polynomial for x = What is the numerical value of a polynomial for x = a?a?

What is a root of a polynomial?What is a root of a polynomial? What is a factor of a polynomial?What is a factor of a polynomial? What is the relationship between roots and factors?What is the relationship between roots and factors? How can we determine the roots?How can we determine the roots? How can we factor a polynomial?How can we factor a polynomial?

Numerical value of a polynomial for x = aNumerical value of a polynomial for x = a

If you evaluate P(x) for x = a, you get If you evaluate P(x) for x = a, you get P(a).P(a).

We say that P(a) is the numerical value We say that P(a) is the numerical value of the polynomial P(x) when x = aof the polynomial P(x) when x = a

Numerical value of a polynomial for x = aNumerical value of a polynomial for x = a

Example: Example: Evaluate P(x)= -2xEvaluate P(x)= -2x22+3x-5 for x +3x-5 for x

= -3= -3

P(-3)= -2(-3)P(-3)= -2(-3)22 + 3(-3) – 5 = + 3(-3) – 5 =325992

So, the numerical value of the polynomial P(x) when x = -3 is: P(-3) = -32

Root of a polynomial Root of a polynomial P(x)P(x)

We say that the number “a” is a root of the We say that the number “a” is a root of the polynomial P(x) if the numerical value polynomial P(x) if the numerical value P(a)=0P(a)=0

In other words: In other words: If, when you evaluate P(x) for x = a, you If, when you evaluate P(x) for x = a, you get P(a)=0, then “a” is a root of P(x)get P(a)=0, then “a” is a root of P(x)

Check that a = 2 and b = 1, are Check that a = 2 and b = 1, are both roots of the polynomial:both roots of the polynomial:

P(x) = x2 - 3x + 2

but c = -1 is not. but c = -1 is not.

Exercise:Exercise:

Factors of a Factors of a polynomialpolynomial We can write the We can write the

polynomial P(x) this way:polynomial P(x) this way: P(x) = x2 - 3x + 2 or this way:or this way: P(x) = (x – 2)(x – 1)

The The linear binomialslinear binomials (x-2) y (x-1) are the (x-2) y (x-1) are the simplest factorssimplest factors of P(x) of P(x)

Searching for rootsSearching for roots

We have to find all numbers We have to find all numbers “a” so that P(a) = 0“a” so that P(a) = 0

But, how will we find them?But, how will we find them?

There are several There are several key ideaskey ideas which will help us to which will help us to find roots:find roots:

The remainder theoremThe remainder theorem

The Ruffini’s ruleThe Ruffini’s rule

This useful property of polynomials:This useful property of polynomials:

““An integer root of a polynomial, is always a divisor of its An integer root of a polynomial, is always a divisor of its constant term” constant term”

The remainder theoremThe remainder theorem :: "If a polynomial P(x) is divided by a binomial (x - a), "If a polynomial P(x) is divided by a binomial (x - a),

then the remainder is equal to the numerical value of then the remainder is equal to the numerical value of the polynomial when x = "a", and can be expressed as the polynomial when x = "a", and can be expressed as P(a)"P(a)"

In other words, we can write:In other words, we can write:

P(x) P(x) x-ax-aRemainder C(x)Remainder C(x)

Remaider = P(a) Remaider = P(a) (numerical value when x = a)(numerical value when x = a)

The remainder theoremThe remainder theorem :: "If a polynomial P(x) is divided by a binomial (x - a), "If a polynomial P(x) is divided by a binomial (x - a),

then the remainder is equal to the numerical value of then the remainder is equal to the numerical value of the polynomial when x = "a", and can be expressed as the polynomial when x = "a", and can be expressed as P(a)"P(a)"

In other words, we can write:In other words, we can write:

P(x) P(x) x-ax-aRemainder C(x)Remainder C(x)

Remaider = P(a) Remaider = P(a) (numerical value when x = a)(numerical value when x = a)

So, if the remainder of the division is zero, then P(a)=0 and therefore “a” is a root of P(x).

The Ruffini’s ruleThe Ruffini’s rule :: This rule is a practical way This rule is a practical way for doing divisions like for doing divisions like

this:this:(x2 – 3x + 2) : (x – 2)

The Ruffini’s ruleThe Ruffini’s rule :: This rule is a practical way This rule is a practical way for doing divisions like for doing divisions like

this:this:(x2 – 3x + 2) : (x – 2)

How?

2

1 -3 2

(x2 – 3x + 2) : (x – 2)

We write this number here

2

1 -3 2

(x2 – 3x + 2) : (x – 2)

We can see that the remainder of the division is We can see that the remainder of the division is zero.zero.

2

1 -3 2

1 -1 0

2 -2

As the remainder of P(x) : (x-2) is zero, “2” is a root of P(x), and therefore, the linear binomial (x-2) is a factor of the polynomial P(x)

Now, we will look for another root in order to get a new factor, and so on… (x2 – 3x + 2) : (x – 2)

This rule makes it easier to find the values of "a".

““An integer root of a polynomial, is always a divisor of its An integer root of a polynomial, is always a divisor of its constant term” constant term”

““An integer root of a polynomial, is always a divisor of its An integer root of a polynomial, is always a divisor of its constant term” constant term”

Example:Example:Which Which integer numbersinteger numbers could becould be roots of the polynomial roots of the polynomialP(x) = - 3x5 + 4x2 – 5x -3 ? ?

Answer: The divisors of the Answer: The divisors of the constant termconstant termThey are only +1, -1, +3 y -3They are only +1, -1, +3 y -3

It is time to use Ruffini’s rule to make the following divisions faster:

P(x):(x-1) P(x):(x+1) P(x):(x-3) and P(x):(x+3)

( x-(-1)) ( x-(3))

So, if we want to factor a polynomial, we need So, if we want to factor a polynomial, we need to find its roots…to find its roots…

Roots2, -1 y 3

1st Factor:(x-2)

2nd Factor:(x+1)

3th Factor:(x-3)

Exercise: Factor the polynomialExercise: Factor the polynomial

Possible integer roots : +1, -1, +2, -2, +3, -3, +6, -6

P(x) = x3 -4x2 + x + 6

Possible integer rootPossible integer root Is it a root? Yes/NoIs it a root? Yes/No FactorFactor

+1+1 NoNo

-1-1 YesYes (x+1)(x+1)

+2+2 YesYes (x-2)(x-2)

-2-2 NoNo

+3+3 YesYes (x-3)(x-3)

-3-3 NoNo

+6+6 NoNo

-6-6 NoNo

Now, we can write the polynomial: Now, we can write the polynomial: P(x) = x3 -4x2 + x + 6

in its factored formin its factored form

P(x) = x3 -4x2 + x + 6 = (x+1)(x-2)(x-3)

In short, there are three strategies for finding In short, there are three strategies for finding roots…roots…

If we “suspect” that a number “a” is a root of the If we “suspect” that a number “a” is a root of the polynomial P(x), we can:polynomial P(x), we can:

1.1. We can make the division P(x) : (x-a). We can make the division P(x) : (x-a). If the remainder is zero, then “a” is a root of P(x).If the remainder is zero, then “a” is a root of P(x).

Ruffini’s ruleRuffini’s rule

2.We can write the number “a” instead of the “x” 2.We can write the number “a” instead of the “x” and make the calculations. If the numerical value, and make the calculations. If the numerical value, P(a)=0, then “a” is a root of P(x).P(a)=0, then “a” is a root of P(x).

Root of a polynomial: definitionRoot of a polynomial: definition

3. We can solve the equation P(x)=03. We can solve the equation P(x)=0Particularly if P(x)=0 is a quadratic equationParticularly if P(x)=0 is a quadratic equation

In a diagram form, we can say:In a diagram form, we can say:

P(x)

Possible integer roots:Divisors of the constant term:

a, b, c…

a is a root b is not a root

(x – a) is a factor

P(x)

Possible integer roots:Divisors of the constant term:

a, b, c…

a is a root b is not a root

(x – a) is a factor

If the remainder of the division is zero…

P(x) x-a 0 C(x)

When the remainder of the division is zero…

P(x) x-a 0 C(x)

P(x)

Possible integer roots:

Divisors of the constant term: a, b, c…

a is a root b is not a root

(x – a) is a factor

If the numerical value P(a) = 0

P(a) = 0

P(x)

Possible integer roots:

Divisors of the constant term: a, b, c…

a is a root b is not a root

(x – a) is a factor

If “a” is a solution of the equation P(x) = 0

Therefore, if we know every root of the polynomial P(x), so to speak, a, b, c, etc…

The factors will be:(x-a), (x-b), (x-c), etc…