We know three of the factors of the original polynomial. ![]() We also need to eliminate the fractions from the factors corresponding to the other two roots. In our case, this LCM is simply 2, and our quadratic becomes:Ģx 2 - 6x + 3. So let's multiply all the coefficients of our quadratic by the lowest common multiple of the denominators of its coefficients. Now we know that this cannot be right, as all of the coefficients of the original polynomial are integers. So our quadratic term looks like x 2 - 3x + 3/2. Let's assume that this factor is the product of the two linear factors corresponding to these roots, (x-a)(x-b). They would seem to be the two roots of a quadratic factor. The other two roots are not recognisable as anything simple. It seems that two of the roots are -4/3 and 3/2. I opened up my trusty UltimaCalc calculator, set it to 12 digit precision, and asked it to try finding the roots, using its built-in polynomial roots calculator. As the degree is even, we are not guaranteed to be able to find any roots (there might not be any!.Īt this point, unless we have a great stroke of luck, we need help. Notice that the polynomial has degree 6, an even number. OK, can we find some roots and divide out the corresponding factors? Yes, if we can find any. But only if we were desperate! There are just too many possibilities. We could look for trial factors based on the possible factors of the coefficients 36 and -72. Simple enough? OK, let's consider factorising this one:ģ6x 6 - 66x 5 - 128x 4 + 131x 3 + 168x 2 - 6x - 72 Try using the values -1 and -3 for x in the original polynomial. The trial and error approach quickly shows that this can be factorised as (x+3)(x+1). We can divide x 3 + 2x 2 - 5x - 6 by x-2 to obtain the quadratic polynomial x^2 + 4x + 3. Looking at the first example, if we try out a few values of x we quickly discover that the polynomial is 0 when x = 2. If we could find a root of a polynomial, we would then be able to divide it by the corresponding factor and end up with a simpler polynomial to work with. Both of these example polynomials are of odd order, so we know that they have at least one root. Perhaps another approach might be simpler. The answer is given at the bottom of this page, here. I challenge you to have a go at finding this solution. Still, the solution can be found, if we have enough patience. The values of b, d and f can come from a set of consisting of 24 different values, as we need to consider negative numbers as well as positive ones. We now have too many choices open to us.įor example, a, c and e have values taken from the set 1, 2, 3, 4, 6, or 12. ![]() Where we know that a, c and e are factors of 12 and that b, d and f are factors of -60. We would now have to consider a factorisation into: Besides, our guess at the form of the factorisation happened to be correct - we do have three linear factors.īut what if we want to factorise something like: After all, we do not have so many choices to try out, here. We can figure out the correct set of values by trial and error. This means that we need to also consider the values -1, -2, -3 and -6. This gives us a clue that the values of a, b and c need to be chosen from the set 1, 2, 3 and 6. In this example, it is clear that the product of a, b and c is -6. This expression can be multiplied out to obtain: For the example above, let's assume that it can be factorised into the form: Whether or not this is possible in theory, the factorisation problem can often seem insurmountable in practice. On the other hand, when the degree is even, there is no guarantee that the polynomial will be zero for any value of x.įactorising has the aim of rewriting a polynomial as a product of some lower order polynomials. Therefore, at some point in between the polynomial must pass through zero at least once. The two values of the polynomial will be of opposite signs. To see this, consider the situations where x is very large and negative, and where x is very large and positive. (The degree of a polynomial is the greatest power of its variable, most often x.) A polynomial whose degree is odd will certainly evaluate to zero for at least one value of x. The example polynomial has a degree of 3. ![]() We can ask: For what value (or values) of x, if any, does the polynomial evaluate to zero? And can the polynomial be expressed as the product of simpler polynomials? Consider for example the following polynomial:
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