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Core 1 Maths problem. Can someone help me understand? Watch

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    Hi, this question is on polynomials revision booklet.
    It says:

    You are given that f(x) = x^3 + 9x^2 + 20x + 12

    i) Show that x=-2 is a root of f(x) = 0

    My question is, how do I show that something is a root?
    Do I just sub f(2) into the equation and get 2^3 + 9(2^2) + 20(2) + 12 = 0
    ?
    but then what would be the next step to show its a root?

    I'm getting confused between the terms "factor" and "root" what is the difference?

    Please help clear it up for me
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    (Original post by wednesday_adams)
    Hi, this question is on polynomials revision booklet.
    It says:

    You are given that f(x) = x^3 + 9x^2 + 20x + 12

    i) Show that x=-2 is a root of f(x) = 0

    My question is, how do I show that something is a root?
    Do I just sub f(2) into the equation and get 2^3 + 9(2^2) + 20(2) + 12 = 0
    ?
    but then what would be the next step to show its a root?

    I'm getting confused between the terms "factor" and "root" what is the difference?

    Please help clear it up for me
    Yes you just plug in x=-2 and show that it is zero as you have done. A root is x=-2 in this case. A factor is (x+2) as if a is a root of f(x)=0 then (x-a) is a factor of f(x).

    Consider that you can write f(x) as (x-a)(x-b)(x-c) if f(x) has 3 roots so each of (x-a),(x-b) and (x-c) is a factor of f(x) since some multiple of it is equal to f(x)
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    Once you have made the substitution and got f(-2) = 0 then you mst state that therefore x = -2 is a root (or solution). Just substituting on its own is not enough.
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    "..therefore, by the factor theorem, (x-2) is a root of the given polynomial..."
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    (Original post by alex2100x)
    Yes you just plug in x=-2 and show that it is zero as you have done. A root is x=-2 in this case. A factor is (x+2) as if a is a root of f(x)=0 then (x-a) is a factor of f(x).

    Consider that you can write f(x) as (x-a)(x-b)(x-c) if f(x) has 3 roots so each of (x-a),(x-b) and (x-c) is a factor of f(x) since some multiple of it is equal to f(x)
    thank you! I understand where I went wrong now. I was plugging in f(2) rather than -2.

    But part 2 says:

    (ii) Divide f(x) by x+6

    and I have no idea how to do this..

    do I plug f(-6) in or something? but how?
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    You should be able to factorise f(x) into brackets. One of those brackets should be (x-6) and therefor you can cancel the two brackets out


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    I should add... Add c1 if youre asked to factorise a cubic, just take an x out and factorise the quadratic


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    (Original post by Factorisation)
    I should add... Add c1 if youre asked to factorise a cubic, just take an x out and factorise the quadratic


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    Taking the x out will not help him as there is a term independent of x, namely the +12. See my other post on polynomial long division on the other thread OP.
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    (Original post by Factorisation)
    You should be able to factorise f(x) into brackets. One of those brackets should be (x-6) and therefor you can cancel the two brackets out


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    He is dividing by (x+6) so we should expect (x+6) to be a factor of f(x) not (x-6)
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    (Original post by alex2100x)
    He is dividing by (x+6) so we should expect (x+6) to be a factor of f(x) not (x-6)
    Oops sorry, my bad!


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    (Original post by Hasufel)
    "..therefore, by the factor theorem, (x-2) is a root of the given polynomial..."
    be careful....

    the root is -2

    the factor is ( x + 2 )
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    (Original post by the bear)
    be careful....

    the root is -2

    the factor is ( x + 2 )
    s**t! - my bad! -i hate myself for that! - sorry fellow posters!

    :oops:
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    (Original post by Hasufel)
    s**t! - my bad! -i hate myself for that! - sorry fellow posters!

    :oops:
    :spank:
 
 
 
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