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Math Help - simplifying expressions

  1. #1
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    simplifying expressions

    can someone please help me simplify these three expressions if possible
    ...and if possible please show working out so i can understand better thanks

    (a) (x+1)^3 -3x (x+1)

    (b) y(2y+3) +2y^2 (y+1)(y-1)

    (c) 14x^3 (3-2x) / x^2(x-1)(4x-6)
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  2. #2
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    Quote Originally Posted by raza88sab View Post
    (a) (x+1)^3 -3x (x+1)
    This is similar in concept to
    a^3 - ba = a(a^2 - b)

    Here a = x + 1 and b = 3x

    So
    (x + 1)^3 - 3x(x + 1) = (x + 1) ( (x + 1)^2 - 3x)

    = (x + 1)(x^2 + 2x + 1 - 3x) = (x + 1)(x^2 -x + 1)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by raza88sab View Post
    (b) y(2y+3) +2y^2 (y+1)(y-1)
    The only thing I can think of to do here is expand:
    y(2y + 3) + 2y^2(y + 1)(y - 1)

    = 2y^2 + 3y + 2y^2(y^2 - 1)

    = 2y^2 + 3y + 2y^4 - 2y^2

    = 2y^4 + 3y

    = y(2y^3 + 3)

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by raza88sab View Post
    (c) 14x^3 (3-2x) / x^2(x-1)(4x-6)
    \frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)}

    Note that 4x - 6 = 2(2x - 3) = -2(3 - 2x)

    So
    \frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)} = -\frac{14x^3(3 - 2x)}{2x^2(x - 1)(3 - 2x)}

    Now we may cancel the common 3 - 2x and the common 2x^2, provided we remember that x \neq 0, \frac{3}{2}.
    \frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)} = -\frac{14x^3(3 - 2x)}{2x^2(x - 1)(3 - 2x)} = -\frac{7x}{x - 1};~x \neq 0, \frac{3}{2}

    -Dan
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  5. #5
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    thanks

    thanks dan mate....im just going to see if i can do this my self and lets see if it sticks in my head somehere lol.....much appriciated
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