Select the polynomial that can be factored as (4x-5)(4x+5)

answer: grouping.

example 1: factoring 2x^2+8x+3x+122x

2

+8x+3x+122, x, squared, plus, 8, x, plus, 3, x, plus, 12

first, notice that there is no factor common to all terms in 2x^2+8x+3x+122x

2

+8x+3x+122, x, squared, plus, 8, x, plus, 3, x, plus, 12. however, if we group the first two terms together and the last two terms together, each group has its own gcf, or greatest common factor:

in particular, there is a gcf of 2x2x2, x in the first grouping and a gcf of 333 in the second grouping. we can factor these out to obtain the following expression:

2x(x+4)+3(x+4)2x(x+4)+3(x+4)2, x, left parenthesis, x, plus, 4, right parenthesis, plus, 3, left parenthesis, x, plus, 4, right parenthesis

how did you do that?

\teald{2x}start color #01a995, 2, x, end color #01a9952x^2+8x

2, x, squared, plus, 8, x\teald{2x}start color #01a995, 2, x, end color #01a995

\dfrac{2x^2}{\teald{2x}}=x

start fraction, 2, x, squared, divided by, start color #01a995, 2, x, end color #01a995, end fraction, equals, x

\dfrac{8x}{\teald{2x}}=4

start fraction, 8, x, divided by, start color #01a995, 2, x, end color #01a995, end fraction, equals, 4

2x^2+8x=\teald{2x}(x+4)

2, x, squared, plus, 8, x, equals, start color #01a995, 2, x, end color #01a995, left parenthesis, x, plus, 4, right parenthesis

\purplec{3}start color #aa87ff, 3, end color #aa87ff3x+123, x, plus, 12\purplec{3}start color #aa87ff, 3, end color #aa87ff

\dfrac{3x}{\purplec{3}}=x

start fraction, 3, x, divided by, start color #aa87ff, 3, end color #aa87ff, end fraction, equals, x

\dfrac{12}{\purplec {3}}=4

start fraction, 12, divided by, start color #aa87ff, 3, end color #aa87ff, end fraction, equals, 4

3x+12=\purplec{3}(x+4)3, x, plus, 12, equals, start color #aa87ff, 3, end color #aa87ff, left parenthesis, x, plus, 4, right parenthesis

2x^2+8x+3x+12=2x(x+4)+3(x+4)

2, x, squared, plus, 8, x, plus, 3, x, plus, 12, equals, 2, x, left parenthesis, x, plus, 4, right parenthesis, plus, 3, left parenthesis, x, plus, 4, right parenthesis

notice that this reveals yet another common factor between the two terms: \goldd{x+4}x+4start color #e07d10, x, plus, 4, end color #e07d10. we can use the distributive property to factor out this common factor.

since the polynomial is now expressed as a product of two binomials, it is in factored form. we can check our work by multiplying and comparing it to the original polynomial. i'd like to see this, !

\begin{aligned}(x+4)(2x+3)& =(x+4)(2x)+(x+4)(3)\\ \\ & =2x^2+8x+3x+12 \end{aligned}

example 2: factoring 3x^2+6x+4x+83x

2

+6x+4x+83, x, squared, plus, 6, x, plus, 4, x, plus, 8

let's summarize what was done above by factoring another polynomial.

\begin{aligned}& \phantom{=}3x^2+6x+4x+8 & =(3x^2+6x)+(4x+8)& & \small{\gray{\text{group terms}}}\\ \\ & =3x({x+2})+4({x+2})& & \small{\gray{\text{factor out gcfs}}}\\ \\ & =3x(\goldd{x+2})+4(\goldd{x+2})& & \small{\gray{\text{common factor! }}} & =(\goldd{x+2})(3x+4)& & \small{\gray{\text{factor out } x+2}} \end{aligned}

​

=3x

2

+6x+4x+8

=(3x

2

+6x)+(4x+8)

=3x(x+2)+4(x+2)

=3x(x+2)+4(x+2)

=(x+2)(3x+4)

​

​

group terms

factor out gcfs

common factor!

factor out x+2

​

the factored form is (x+2)(3x+4)(x+2)(3x+4)left parenthesis, x, plus, 2, right parenthesis, left parenthesis, 3, x, plus, 4, right parenthesis.

look a the sides they are perpendicular

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above