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anonymous

  • one year ago

Explain how a four-term polynomial is factored by grouping and when a quadratic trinomial can be factored using this method.

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  1. anonymous
    • one year ago
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    soo what do u know about each of them

  2. anonymous
    • one year ago
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    So say for example we have x5−3x3−2x2+6 We group the terms together by 2's (x5−3x3)−(2x2−6) Notice how that highlighted '-' is there...it WAS '+' but since we have a '-' in front of the parenthesis...when that distributes..it turns that '+' '-' again So now...we factor out of each parenthesis what we can... x^5 - 3x^3 ...well looks like at least x^3 can come out of there 2x^2 - 6 ...well we can factor a 2 out of there.. so now we have x3(x2−3)−2(x2−3) Now notice how we have 2 parenthesis that have the same thing inside them...and we also have 2 coefficients of those parenthesis...so we put those together respectively (coefficients in here)(those things in the parentheis since they are the same) (x3−2)(x2−3) Everything make sense?

  3. anonymous
    • one year ago
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    Well say we have 3x2+10x−8 We can actually break up that 10x ...and make it -2x + 12x right? since -2 + 12 =10 so we would then have 3x2−2x+12x−8 NOW we can group these up by 2's (3x2−2x)+(12x−8) Lets factor out an 'x' of the first set of parenthesis...and also lets factor out a 4 of the last set... x(3x−2)+4(3x−2) Look the same thing as before...same thing in both parenthesis.and 2 coefficients outside...so again (x+4)(3x−2)

  4. anonymous
    • one year ago
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    2nd comment is for trinomial

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