# how can I factorise this cubic

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

Then employ the factor theorem and long division.

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(Original post by

Can you guess a root?

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

Then employ the factor theorem and long division.

**RDKGames**)Can you guess a root?

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

Then employ the factor theorem and long division.

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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(Original post by

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?

**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?

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 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 , then by the Factor Theorem we know that is a factor of the cubic.

This means we can use long division to divide the cubic by and end up with a quadratic which is more obvious to solve and obtain the other two roots out of.

Last edited by RDKGames; 1 year ago

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**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

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....

**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....

but i hate this method it makes me doubt everything I'm doing. Is there another way?

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(Original post by

Let's take a quadratic equation case for example. Suppose we know two roots of it, and call them . Then we know we can factorise this quadratic into the form . But now if we expand the LHS we get . Notice that the constant term is precisely the product 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 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 , then by the Factor Theorem we know that is a factor of the cubic.

This means we can use long division to divide the cubic by and end up with a quadratic which is more obvious to solve and obtain the other two roots out of.

**RDKGames**)Let's take a quadratic equation case for example. Suppose we know two roots of it, and call them . Then we know we can factorise this quadratic into the form . But now if we expand the LHS we get . Notice that the constant term is precisely the product 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 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 , then by the Factor Theorem we know that is a factor of the cubic.

This means we can use long division to divide the cubic by and end up with a quadratic which is more obvious to solve and obtain the other two roots out of.

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(Original post by

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?

**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?

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