Factorise the following p(x) as the product of a linear term and a quadratic polynomial with no real roots.

Question

Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them): 1. $$p(x)=x^3+x$$ 2. $$p(x)=x^4+x^2$$ 3. $$p(x)=x^3-x$$

Answer 1

Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.

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1. $$x^3+x = x(x^2+1)$$ Roots are $0$, $i$, and $-i$.
2. $$x^4+x^2 = x^2(x^2+1)$$ Roots are $0$, $i$, and $-i$.
3. $$x^3-x = x(x^2-1) = x(x+1)(x-1)$$ Roots are $0$, $1$, and $-1$.

Answer 2

I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.

We have that if for some polynomial $p(x)$, $p(a)=0$, then:

$$p(x)=(x-a)\cdot q(x)$$

Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).

If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.

Let me demonstrate first on (b)

We have that $p(x)=x^4-x^2$ I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:

$$x^4-x^2=x\cdot q(x)$$

From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2\cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.

This is a rather bland example, so I'll demonstrate the procedure on a better example.

Suppose we have $$f(x)=x^3+4x^2+7x+4$$

Plugging in values for $x$, we get that $f(-1)=0$

So we have that:

$$x^3+4x^2+7x+4=(x+1)\cdot g(x)$$ Or rather,

$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.

Expanding this out we get:

$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$

From which it follows that $p=3, q=4$.

So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)