# We have to solve the following equation (the matix in the document). How am i meant to figure out what a=? i have done some working but i am not sure what ot do next. i am not sure what to do...

We have to solve the following equation (the matix in the document).

How am i meant to figure out what a=?

i have done some working but i am not sure what ot do next.

i am not sure what to do with the 9. I have factorised the forst part.

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You have correctly found the determinant to be `x^3-5x^2+3x+9`

Since the matrix is singular the determinant is zero so:

`x^3-5x^2+3x+9=0`

There are a number of ways to try to factor this. The only possible rational roots are `+-1,+-3` so it doesn't hurt to try them. You will find that 3 is a root so (x-3) is a factor.

`(x-3)(x-3)(x+1)=0`

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The two solutions are x=3 (double root) and x=-1

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You can also use graphing technology to check for the roots. Real roots will be the x-intercepts of the graph of `y=x^3-5x^2+3x+9` .

how do we know that the only possible rational roots are +-1 and +-3?

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For a polynomial of the form `a_nx^n+a_(n-1)x^(n-1)+...+a_1x+a_0` , the rational root theorem guarantees that the only possible rational roots, roots of the form `p/q` , are rational numbers such that p is a factor of the constant term `a_0` and q is a factor of the leading coefficient `a_n` .

Hi

can you please explain how you got

and why do we take 1 and 3 as possible roots?

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I used synthetic division, but you could guess at a possible factor and use long division, guess a possible root and use the remainder theorem to check if it is a root, or you could use a graphing utility to find the zeros.

Synthetic division: Guess 3 as a root. (Why? Why not? The only possible roots are of teh form `p/q` where p is a factor of 9 and q is a factor of 1 so you need only check `+-1,+-3,+-9` . You might methodically check each one until you find a root.)

3 | 1 -5 3 9

3 -6 -9

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1 -2 -3 0

So (x-3) is a factor and the other factor is `x^2-2x-3` , which factors further. (Now you can use the quadratic theorem if you like.)