Factoring $f(x)=3x^4+14x^3+21x^2+10x$

Question

How should I go about factoring $f(x)=3x^4+14x^3+21x^2+10x$

I tried factoring out a $x$, and then factoring by grouping, but that led me to no avail.

I used a polynomial calculator and it gave me the answer without any of the steps as to how they did it.

Any suggestions?

Answer 1

HINT

Notice that $f(0) = f(-1) = 0$. Taking advantage of such facts, one has \begin{align*} f(x) & = 3x^{4} + 14x^{3} + 21x^{2} + 10x\\\\ & = x(3x^{3} + 14x^{2} + 21x +10)\\\\ & = x[(3x^{3} + 3x^{2}) + (11x^{2} + 11x) + (10x + 10)]\\\\ & = x[3x^{2}(x+1) + 11x(x+1) + 10(x+1)]\\\\ & = x(x+1)(3x^{2} + 11x + 10) \end{align*}

Can you take it from here?

Answer 2

There is a theorem called the rational root theorem which states that all rational roots of an integer polynomial are of the form $\frac{p}{q}$ where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient, when neither are equal to $0$. You already got that $$f(x) = x(3x^3 + 14x^2 + 21x + 10)$$

Then the rational roots of the second term must be in $\left\{ \pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{5}{3}, \pm \frac{10}{3} \right\}$. Once you go through this list and find a root $r$, you can factor out $x-r$. For example, with the root $x=-2$, the equation becomes $$f(x)=x(x+2)(3x^2+8x+5)$$

The last term can be factored, with the quadratic formula, by simply recognizing that $3+5 = 8$ and thus $x+1$ would be a factor, or by continuing to go through the possible list of rational roots. Ultimately, it would factor out to $$f(x)=x(x+2)(x+1)(3x+5)$$

Answer 3

Answer : f(-1)=0⇒f(x) = (x+1)(3x³+11x²+10x) ⇒f(x) = x(x+1)(3x^2 +11x+10)

On pose h(x) = 3x^2 +11x+10

⇒f(x) =x(x+1)(h(x))

h(-2)=0⇒ h(x) =(x+2)(3x+5)

Finally : ⇒ f(x) = x(x+1)(x+2)(3x+5)

Answer 4

Wolfram Alpha can often show you factors in alternate forms as shown here about a third of the way down the page.

Another way of finding $1$ of the solutions to a cubic equation is to use the cubic formula;

$$x=\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}+\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a}$$

where, for $\quad(3x^3 + 14x^2 + 21x + 10)\qquad a=3\quad b=14\quad c=21\quad d=10$ \begin{equation} x=\sqrt[3]{\biggl(\frac{-(14)^3}{27(3)^3 }+\frac{(14)(21)}{6(3)^2}-\frac{(10)}{2(3)}\biggr)+\sqrt{\biggl(\frac{-(14)^3}{27(3)^3}+\frac{(14)(21)}{6(3)^2}-\frac{(10)}{2(3)}\biggr)^2+\biggl(\frac{(21)}{3(3)}-\frac{(14)^2}{9(3)^2}\biggr)^3}}+\sqrt[3]{\biggl(\frac{-(14)^3}{27(3)^3 }+\frac{(14)(21)}{6(3)^2}-\frac{(10)}{2(3)}\biggr)-\sqrt{\biggl(\frac{-(14)^3}{27(3)^3}+\frac{(14)(21)}{6(3)^2}-\frac{(10)}{2(3)}\biggr)^2+\biggl(\frac{(21)}{3(3)}-\frac{(14)^2}{9(3)^2}\biggr)^3}}-\frac{(14)}{3(3)} \end{equation}

I haven't tried this in Excel or with Wolfram Alpha because it takes to loog to remove the things \frac{}{}, \bigg, \sqrt[]{}, etc that go into making it. It has worked when I plugged in numbers from other equations even when I had transient complex numbers where I needed robust software like Wolfram Alpha to resolve them.