username2855962
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Would someone help me understand the method to this question

The cubic polynomial f(x) is defined by f(x)=x^3+x^2−11x+10

It can be shown that (x−2) is a factor of f(x).

Hence solve the equation f(x)=0, giving each root in an exact form.
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RDKGames
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(Original post by The Big "R")
Would someone help me understand the method to this question

The cubic polynomial f(x) is defined by f(x)=x^3+x^2−11x+10

It can be shown that (x−2) is a factor of f(x).

Hence solve the equation f(x)=0, giving each root in an exact form.
(6 marks)
(x-2) is a factor... so, by the Factor Theorem, one of the roots is...??

Now use polynomial division to divide x^3 + x^2 - 11x + 10 by x-2 to obtain a quadratic. Solving this quadratic will yield the other two roots.
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mqb2766
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If you want to show it's a factor, sub x=2 in and show f(2)=0.
Otherwise, divide the cubic by x-2. Then the other two factors are given by factorising/solving the quadratic quotient.
(Original post by The Big )
Would someone help me understand the method to this question

The cubic polynomial f(x) is defined by f(x)=x^3+x^2−11x+10

It can be shown that (x−2) is a factor of f(x).

Hence solve the equation f(x)=0, giving each root in an exact form.
(6 marks)
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username2855962
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(Original post by RDKGames)
(x-2) is a factor... so, by the Factor Theorem, one of the roots is...??

Now use polynomial division to divide x^3 + x^2 - 11x + 10 by x-2 to obtain a quadratic. Solving this quadratic will yield the other two roots.
The quadratic i got was x^2 + 3x -5 when dividing and i get -3 + and - root29 divided by 2. However i dont understand what you mean by the first point x-2 is a factor?
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username2855962
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(Original post by RDKGames)
(x-2) is a factor... so, by the Factor Theorem, one of the roots is...??

Now use polynomial division to divide x^3 + x^2 - 11x + 10 by x-2 to obtain a quadratic. Solving this quadratic will yield the other two roots.
Dont worry ive got it now thanks for your help
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RDKGames
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(Original post by The Big "R")
The quadratic i got was x^2 + 3x -5 when dividing and i get -3 + and - root29 divided by 2. However i dont understand what you mean by the first point x-2 is a factor?
(x-2) being a factor of a polynomial R(x) means this polynomial can be rewritten as R(x) = (x-2)Q(x) where Q(x) is a different polynomial.

The solutions to R(x) = 0 are given by x-2 = 0 or Q(x) = 0. Clearly, x=2 is a solution to the first factor. To solve Q(x) = 0 you first need to know what Q(x) even is, which is obtained by

R(x) = (x-2)Q(x) \implies \dfrac{R(x)}{x-2} = Q(x).

This is why you need to divide R(x) by x-2.


The Factor Theorem, which you should probably be aware of if you're doing a question like this, says that if (x-a) is a factor of a polynomial R(x), then we must have R(a) = 0, meaning x=a is a root.
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