2.3 Factoring Polynomials Math Worksheet (Page 3 of 7) in pdf

2.3 Factoring Polynomials Math Worksheet Page 3

I checked the answer by multiplying

Check:

(x 1 5) (x 2 6)

the two factors.

5 (x 1 5)x 2 (x 1 5)6

5 x

1 5x 2 6x 2 30

5 x

2 x 2 30

First I divided each term by the

18x

2 50

common factor, 2. This left a

5 2(9x

2 25)

difference of squares.

I took the square root of

and 25

5 2(3x 1 5) (3x 2 5)

to get

and 5, respectively.

This is a trinomial of the form

10x

2 x 2 3

1 bx 1 c,

where

a 2 1,

and it has

no common factor.

I used decomposition by finding two

5 10x

1 5x 2 6x 2 3

numbers whose sum is

and whose

product is

These

(10) (23) 5 230.

numbers are 5 and

I used them to

26.

“decompose” the middle term.

I factored the group consisting of the

5 5x(2x 1 1) 2 3(2x 1 1)

first two terms and the group

consisting of the last two terms by

dividing each group by its common

factor.

I divided out the common factor of

5 (2x 1 1) (5x 2 3)

2x 1 1 from each term.

I noticed that the first and last terms

1 30x 1 25

are perfect squares. The square roots

are

and 5, respectively. The middle

5 (3x 1 5)

term is double the product of the two

square roots,

2(3x) (5) 5 30x.

So this

trinomial is a perfect square, namely,

the square of a binomial.

Trinomials of this form may be

e) 2x

1 x 1 3

factored by decomposition. I tried to

come up with two integers whose

sum is 1 and whose product is 6.

There were no such integers, so the

trinomial cannot be factored.

100

2.3 Factoring Polynomials

2.3 Factoring Polynomials Math Worksheet Page 3

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