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    Given that f(x) = x^3 = 2x^2 - 5x - 6 use the Factor Theorem to factorise f(x) completely, hence solve the inequality x^3 + 2x^2 - 5x - 6 > 0
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    (Original post by Sniper21)
    Given that f(x) = x^3 = 2x^2 - 5x - 6 use the Factor Theorem to factorise f(x) completely, hence solve the inequality x^3 + 2x^2 - 5x - 6 > 0
    Guess a root and factorise. A good start is to plug in factors of the constant term.
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    Factor theorem says that if f(a)=0 then (x-a) is a factor and since this is a cubic f(x)=(x-a)(bx^2+cx+d)=x^3+2x^2-5x-6.Find a suitable a(try factors of -6) and then equate coefficients to find b,c and d and then factorise the quadratic.
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    (Original post by Sniper21)
    Given that f(x) = x^3 = 2x^2 - 5x - 6 use the Factor Theorem to factorise f(x) completely, hence solve the inequality x^3 + 2x^2 - 5x - 6 > 0
    Is that all the information that is given?
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    (Original post by Sniper21)
    Given that f(x) = x^3 = 2x^2 - 5x - 6 use the Factor Theorem to factorise f(x) completely, hence solve the inequality x^3 + 2x^2 - 5x - 6 > 0
    I'm doing additional maths gcse this year and have self taught this so I'm gonna give it a go, no guarantees that it will be correct.

    By substituting values in for x I can see that f(2) = 0
    So (x-2) is one of the three brackets
    Then we divide f(x) by (x-2)
    x^3+2x^2-5x-6 = (x-2) + x^2+4x+4
    f(x) = (x-2)(x+1)(x+3)
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    (Original post by RDKGames)
    Guess a root and factorise. A good start is to plug in factors of the constant term.
    This is the first part of the question
    Use the Remainder Theorem to find the remainder when f(x) = 8x^3 - 4x^2 + 6x +7 is divided by x-1

    Please show me the step by step solution.
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    (Original post by leadheadsmith)
    Is that all the information that is given?
    No, this is the original question:
    Use the remainder theorem to find the remainder when f9x) = 8x^3 - 4x^2 +6x+7 is divided by x-1
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    (Original post by Sniper21)
    This is the first part of the question
    Use the Remainder Theorem to find the remainder when f(x) = 8x^3 - 4x^2 + 6x +7 is divided by x-1

    Please show me the step by step solution.
    The remainder of a polynomial f(x) when divided by (x-a) is f(a), so you would replace any x in f(x) with an a. Step by step solutions are not allowed.
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    (Original post by Sniper21)
    This is the first part of the question
    Use the Remainder Theorem to find the remainder when f(x) = 8x^3 - 4x^2 + 6x +7 is divided by x-1

    Please show me the step by step solution.
    These basics should be covered in the C1 book shouldn't they?

    Just plug in x=1 and your output is the remainder.
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    (Original post by Sniper21)
    No, this is the original question:
    Use the remainder theorem to find the remainder when f9x) = 8x^3 - 4x^2 +6x+7 is divided by x-1
    8x^2+4x+10

    x-1 | 8x^3 - 4x^2 +6x+7
    -(8x^3-8x^2)
    4x^2+6x+7
    -(4x^2-4x)
    10x+7
    -(10x-10)
    17

    Remainder = 17

    I think it's right, i find these questions easier written down haha
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    (Original post by RDKGames)
    These basics should be covered in the C1 book shouldn't they?

    Just plug in x=1 and your output is the remainder.
    This is what I got for part a) Remainder theorem:
    polynomial f(x) has remainder r when divided by (x-a) if f(a) = r
    f(x) = 8x^3 - 4x^2 + 6x + 7
    f(x) = 8 - 4 + 6 + 7 = 17
    It is very straightforward, thanks a lot for your help!
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    (Original post by Sniper21)
    This is what I got for part a) Remainder theorem:
    polynomial f(x) has remainder r when divided by (x-a) if f(a) = r
    f(x) = 8x^3 - 4x^2 + 6x + 7
    f(x) = 8 - 4 + 6 + 7 = 17
    It is very straightforward, thanks a lot for your help!
    No worries Don't forget to give us all the information in future though!
 
 
 
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