Factoring Difference of Squares help? 5 problems.?

Can you please do these 5 problems and show me your work? I just am not understanding how to do them.

1) 2p^3 - 162p

2) u^4 - 81

3) 2d^4 - 32f^4

4) 20r^4 - 45n^4

5) 256n^4 - c^4

4 Answers

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  • 8 years ago
    Favorite Answer

    1)2p³ - 162p

    =2p(p² - 81)

    =2p(p² - 9²)

    =2p (p + 9) (p - 9)

    2)u⁴ - 81

    =u⁴ - 3⁴

    =(u²)² - (3²)²

    =(u² + 3²) (u² - 3²)

    =(u² + 9) (u + 3) (u - 3)

    3)2d⁴ - 32f⁴

    =2(d⁴ - 16f⁴)

    =2{ (d²)² - (4f²)² }

    =2 { (d² + (4f)² } { d² - (4f)² }

    =2 (d² + 4f²) ( d + 4f) (d - 4f)

    4)20r⁴ - 45n⁴

    =5(4r⁴ - 9n⁴)

    =5{ (2r²)² - (3n²)² }

    =5 (2r² + 3n²) (2r² - 3n²)

    5)256n⁴ - c⁴

    =(16n²)² - (c²)²

    =(16n² + c²) (16n² - c²)

    =(16n² + c²) { (4n)² - (c)² }

    =(16n² + c²) (4n + c) (4n - c)

  • Colin
    Lv 7
    8 years ago

    Fundamental truth:

    a^2 - b^2 = (a - b)*(a + b)

    3) 2d^4 - 32f^4

    = 2*(d^4 - 16*f^4)

    Recognise as difference of squares

    = 2*([d^2]^2 - [4*f^2]^2)

    = 2*(d^2 -4*f^2)*(d^2 +4*f^2)

    First term is difference of squares again

    = 2*(d - 2*f)*(d + 2*f)*(d^2 +4*f^2) <<<

    4) 20r^4 - 45n^4

    Take out a factor of 5...

    5) 256n^4 - c^4

    = ([16*n^2]^2 - [c^2]^2)

    and off you go

  • 4 years ago

    (a^2)-4 = 0 (a-2) (a+2) = 0 a-2 = 0 or a+2 = 0 a = 2 or a = -2 a million-49c^2 a million-40 9 c^2 = 0 -((7 c-a million) (7 c+a million)) = 0 multiply the two area cases -a million to do away with the - (7 c-a million) (7 c+a million) = 0 7 c-a million = 0 or 7 c+a million = 0 7 c = a million or 7 c = -a million c = a million/7 or c = -a million/7

  • DWRead
    Lv 7
    8 years ago

    2p³ - 162p = 2p(p² - 81)

         = 2p(p + 9)(p - 9)

    :::::

    u⁴ - 81 = (u²)² - 9²

       = (u² + 9)(u² - 9)

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