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.

ya that's cool thanks for the points