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anonymous

  • 5 years ago

factor th trinomial if possible using integers show work 2x^2+7x+5

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  1. anonymous
    • 5 years ago
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    \[2x ^{2} + 7x + 5\] = (2x + a)(x + b) First Outside Inside Last (FOIL Method) 2x*x + 2x*b + a*x + a*b That gives our \[2x ^{2}\] Then 2x*b + a*x = 7x and a*b = 5 From the second equation, we know that a and b are some combination of 1 and 5 (because these are the only two integers which, when multiplied, equal 5). If we set a=1 and b=5 we have: (2x + 1)(x + 5) Let's see. 2x^2 + 10x + 1x + 5=2x^+11x+5... That doesn't work. Let's try the other way. (2x+5)(x+1) 2x^2+2x + 5x + 5=2x^2+7x+5 Looks good too me! Was that enough work shown?

  2. anonymous
    • 5 years ago
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    Or are you supposed to use the quadratic equation?

  3. anonymous
    • 5 years ago
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    yeah thats good thank you

  4. myininaya
    • 5 years ago
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    a=2 b=7 c=5 find two factors of a*c that add up to be b a*c=2(5) b=7=2+5 now replace middle term the bx term with 2x+5x so we have 2x^2x+2x+5x+5 now factor by grouping

  5. myininaya
    • 5 years ago
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    2x(x+1)+5(x+1) (x+1)(2x+5)

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