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Mathematics 9 Online
OpenStudy (anonymous):

Chase wants to factor x2 + 10x + 25 by grouping; however, Paige says it is a special product and can factor a different way. Using complete sentences, explain and demonstrate how both methods will result in the same factors. @jim_thompson5910 Please help me out

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

@ganeshie8 @dumbcow

OpenStudy (tkhunny):

Chase is miserably misguided. You should not ever expect to factor by grouping. There is limited benefit in FBG. It is a special type. The first and last terms are prefect squares. I wonder if the whole thing could be a perfect square trinomial?

OpenStudy (anonymous):

can you help me solve for this and see how both would lead me to same answer @tkhunny

OpenStudy (anonymous):

@jim_thompson5910 pleaaaaseeee i need help :/

jimthompson5910 (jim_thompson5910):

what two numbers multiply to 25 and add to 10

OpenStudy (anonymous):

5

jimthompson5910 (jim_thompson5910):

5 and what else

OpenStudy (anonymous):

5 and 5

jimthompson5910 (jim_thompson5910):

good 5+5 = 10 5*5 = 25

jimthompson5910 (jim_thompson5910):

this means 10x breaks down into 5x+5x

jimthompson5910 (jim_thompson5910):

x^2 + 10x + 25 turns into x^2 + 5x + 5x + 25

jimthompson5910 (jim_thompson5910):

now factor by grouping

OpenStudy (anonymous):

so how do i answer this question?

jimthompson5910 (jim_thompson5910):

now factor x^2 + 5x + 5x + 25 by grouping

jimthompson5910 (jim_thompson5910):

examples http://www.regentsprep.org/regents/math/algtrig/atv1/revfactorgrouping.htm

OpenStudy (anonymous):

x(x+5)+5(x+5)? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

keep going

OpenStudy (anonymous):

so both the answers have (x+5) (x+5) so they're both identical

jimthompson5910 (jim_thompson5910):

yes, the other way is to use the special formula \[\Large a^2 + 2ab + b^2 = (a+b)^2\] the perfect square trinomial formula

OpenStudy (anonymous):

okay is it okay if we use twiddla for this again?

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