Free Powerpoint Presentations

Factoring Polynomials
Page
5

DOWNLOAD

PREVIEW

WATCH ALL SLIDES

Since the middle term is negative, possible factors of 6 must both be negative: {1,  6} or { 2,  3}.

We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors.

Factoring Polynomials

Example

Continued.

Slide 31

We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 7x.

We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 7x.

{1, 6} (3x 1)(x 6) 18x x 19x

(3x 6)(x 1) Common factor so no need to test.

{2, 3} (3x 2)(x 3) 9x 2x 11x

(3x 3)(x 2) Common factor so no need to test.

Factoring Polynomials

Example Continued

Continued.

Slide 32

Now we have a problem, because we have exhausted all possible choices for the factors, but have not found a pair where the sum of the products of the outside terms and the inside terms is 7.

Now we have a problem, because we have exhausted all possible choices for the factors, but have not found a pair where the sum of the products of the outside terms and the inside terms is 7.

So 3x2 7x + 6 is a prime polynomial and will not factor.

Factoring Polynomials

Example Continued

Slide 33

Factor the polynomial 6x2y2  2xy2  60y2.

Factor the polynomial 6x2y2 2xy2 60y2.

Remember that the larger the coefficient, the greater the probability of having multiple pairs of factors to check. So it is important that you attempt to factor out any common factors first.

6x2y2 2xy2 60y2 = 2y2(3x2 x 30)

The only possible factors for 3 are 1 and 3, so we know that, if we can factor the polynomial further, it will have to look like 2y2(3x )(x ) in factored form.

Factoring Polynomials

Example

Continued.

Slide 34

Since the product of the last two terms of the binomials will have to be 30, we know that they must be different signs.

Since the product of the last two terms of the binomials will have to be 30, we know that they must be different signs.

Possible factors of 30 are {1, 30}, {1, 30}, {2, 15}, {2, 15}, {3, 10}, {3, 10}, {5, 6} or {5, 6}.

We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to x.

Factoring Polynomials

Example Continued

Continued.

Slide 35

{-1, 30} (3x  1)(x + 30) 90x -x 89x

{-1, 30} (3x 1)(x + 30) 90x -x 89x

(3x + 30)(x 1) Common factor so no need to test.

{1, -30} (3x + 1)(x 30) -90x x -89x

(3x 30)(x + 1) Common factor so no need to test.

{-2, 15} (3x 2)(x + 15) 45x -2x 43x

(3x + 15)(x 2) Common factor so no need to test.

{2, -15} (3x + 2)(x 15) -45x 2x -43x

(3x 15)(x + 2) Common factor so no need to test.

Factoring Polynomials

Example Continued

Continued.

Go to page:
 1  2  3  4  5  6  7  8  9  10 

Contents

Last added presentations

2010-2024 powerpoint presentations