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#1
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)
0
1 year ago
#2
(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)
is a factor... so, by the Factor Theorem, one of the roots is...??

Now use polynomial division to divide by to obtain a quadratic. Solving this quadratic will yield the other two roots.
1
1 year ago
#3
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)
1
#4
(Original post by RDKGames)
is a factor... so, by the Factor Theorem, one of the roots is...??

Now use polynomial division to divide by 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?
0
#5
(Original post by RDKGames)
is a factor... so, by the Factor Theorem, one of the roots is...??

Now use polynomial division to divide by to obtain a quadratic. Solving this quadratic will yield the other two roots.
Dont worry ive got it now thanks for your help
0
1 year ago
#6
(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?
being a factor of a polynomial means this polynomial can be rewritten as where is a different polynomial.

The solutions to are given by or . Clearly, is a solution to the first factor. To solve you first need to know what even is, which is obtained by

.

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 is a factor of a polynomial , then we must have , meaning is a root.
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