How do you factorise polynomials by inspection? (1 Viewer)

sadpwner · Jun 18, 2015

eg. Factorise x^4-x^3-2x^2+6x-4 over the real field given that 1+i and 1-i are factors.

InteGrand · Jun 18, 2015

sadpwner said:
eg. Factorise x^4-x^3-2x^2+6x-4 over the real field given that 1+i and 1-i are factors.

$Since $(1+i)$ and $(1-i)$ are factors, we immediately know that one real quadratic factor is $(x^2 - 2x + 2)$. This is because we know that if $(x-\alpha)$ and $(x-\beta)$ are factors, then so is $(x-\alpha)(x-\beta)$, and using our knowledge of sums and products of roots, $(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x+\alpha \beta$, and we can easily find by inspection $\alpha + \beta$ and $\alpha \beta$ in the case given by using the complex number rules $\omega + \overline{\omega}=2\Re (\omega)$ and $\omega \overline{\omega} = |\omega|^2$.$

$Now we seek to find by inspection the other real quadratic factor (clearly this is the one and only other factor as the polynomial has degree 4). To do this, we write $x^4 -x^3 -2x^2 + 6x-4=(x^2 - 2x + 2)(\text{ }\_\_\text{ }x^2 \text{ }\_\_\text{ }x\text{ }\_\_\text{ })$. In other words, leave gaps for the coefficients of the second factor, which we will aim to fill in.$

$To fill these in, we need to find the numbers that would make both sides have the same coefficients for each power of $x$. The easiest coefficients of $x$ to compare are basically all of those except $x^2$, since $x^2$ can be written as a product of powers of $x$ from $x^0$ to $x^2$ (which are those in the quadratic factors) in more ways that the others.$

$We can easily find the missing coefficient of $x^2$: it must be 1, because the LHS has a coefficient of $1$ for its $x^4$ term, and this coefficient is the product of the $x^2$ coefficients in the RHS (since that's the only way to get an $x^4$ term by multiplying out terms in the RHS). So write 1 in the $x^2$ coefficient's blank (or leave it blank if you want, since we don't need to write 1, but it may be better to write in the 1 so you don't get confused about whether you've determined that coefficient yet or not).$

$Similarly, the constant term in the LHS is the product of the constant terms in the quadratic factors, so the missing constant term on the RHS must be $-2$, so fill this in.$

$For the missing $x$ coefficient (call it $b$), by comparing coefficients of $x^3$ on both sides, we want $b$ such that $-1=b-2\Rightarrow b=1$. This is because the ways to get $x^3$ on the RHS are by multiplying the $x^2$ term from one quadratic factor by the $x$ term of the other. This can easily be done by inspection with a little practice. (We could also have found the missing $x$ coefficient by comparing the $x$ coefficients of both sides.)$

$So $x^4 -x^3 -2x^2 + 6x-4=(x^2 - 2x + 2)(1x^2 +1x-2)$ (or $(x^2 - 2x + 2)(x^2 +x-2)$).$

How do you factorise polynomials by inspection? (1 Viewer)

sadpwner

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InteGrand

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