Blonde.
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x^3 - 4x^2 + x + 6

please help i've been trying for ages
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RDKGames
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(Original post by Blonde.)
x^3 - 4x^2 + x + 6

please help i've been trying for ages
Can you guess a root?

Try the factors of 6 (and their -ve versions)

Then employ the factor theorem and long division.
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Blonde.
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(Original post by RDKGames)
Can you guess a root?

Try the factors of 6 (and their -ve versions)

Then employ the factor theorem and long division.
idk where to start I tried guessing but it is wrong

can you explain why it has to be factors of 6, or how you'd guess a root - based on what?, and how the factor theorem and long division comes into factorising this?
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Yr_11_MATHS
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(Original post by Blonde.)
x^3 - 4x^2 + x + 6

please help i've been trying for ages
You have to try many numbers to see if they're a factor... most common ones for A level are 1 (x-1) and 2(x-2)... if not then use your calculator to factorise it for you and then use one of the factors to divide the equation by so that you get a quadratic which can be solved using completing the square....
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RDKGames
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(Original post by Blonde.)
idk where to start I tried guessing but it is wrong

can you explain why it has to be factors of 6, or how you'd guess a root - based on what?, and how the factor theorem and long division comes into factorising this?
Let's take a quadratic equation case for example. Suppose we know two roots of it, and call them \alpha,\beta. Then we know we can factorise this quadratic into the form (x-\alpha)(x-\beta) = 0. But now if we expand the LHS we get x^2 - (\alpha + \beta)x + \alpha\beta = 0. Notice that the constant term is precisely the product \alpha\beta of our roots. This idea extends to any polynomial, so your cubic included.

This means we can look at the constant term 6 and start testing its factors (as well as their -ve versions) to see whether any of them satisfy the equation.

This means you want to test x = \pm 1, \pm 2, \pm 3, \pm 6 by substituting them one at a time into the cubic and seeing if any of them satisfy it.

Once you find one number which does, let's call it a, then by the Factor Theorem we know that x-a is a factor of the cubic.

This means we can use long division to divide the cubic by x-a and end up with a quadratic which is more obvious to solve and obtain the other two roots out of.
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Blonde.
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(Original post by RDKGames)
Can you guess a root?

Try the factors of 6 (and their -ve versions)

Then employ the factor theorem and long division.
(Original post by Yr_11_MATHS)
You have to try many numbers to see if they're a factor... most common ones for A level are 1 (x-1) and 2(x-2)... if not then use your calculator to factorise it for you and then use one of the factors to divide the equation by so that you get a quadratic which can be solved using completing the square....
I worked it out by trial and error to be (x-3)(x-2)(x+1)
but i hate this method it makes me doubt everything I'm doing. Is there another way?
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Blonde.
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(Original post by RDKGames)
Let's take a quadratic equation case for example. Suppose we know two roots of it, and call them \alpha,\beta. Then we know we can factorise this quadratic into the form (x-\alpha)(x-\beta) = 0. But now if we expand the LHS we get x^2 - (\alpha + \beta)x + \alpha\beta = 0. Notice that the constant term is precisely the product \alpha\beta of our roots. This idea extends to any polynomial, so your cubic included.

This means we can look at the constant term 6 and start testing its factors (as well as their -ve versions) to see whether any of them satisfy the equation.

This means you want to test x = \pm 1, \pm 2, \pm 3, \pm 6 by substituting them one at a time into the cubic and seeing if any of them satisfy it.

Once you find one number which does, let's call it a, then by the Factor Theorem we know that x-a is a factor of the cubic.

This means we can use long division to divide the cubic by x-a and end up with a quadratic which is more obvious to solve and obtain the other two roots out of.
thank you so much, that clears up everything wow
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Yr_11_MATHS
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(Original post by Blonde.)
I worked it out by trial and error to be (x-3)(x-2)(x+1)
but i hate this method it makes me doubt everything I'm doing. Is there another way?
The first bit is the trial and error bit, I always use the calculator to tell me a point in which the line intersects the X axis. Then I will normally divide the factor by the equation and solve from there. There's not really a quick way to do it unless you know two roots alpha and beta in which you could do alpha+beta=-b/a and alpha*beta = c/a... but that will be getting into further maths content but still does the trick
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