Can we find the coefficients and factorise this quartic?

Can we find the coefficients and factorise this quartic?

Question

The expression $2x^4+ax^3+bx^2-4x-4$ where $a$ and $b$ are constants, is denoted by $f(x)$. Given that $f\left(-\dfrac{1}{2}\right)=0$ and $f(2)=0$, find the values of $a$ and $b$.

With these values for $a$ and $b$

1. express $f(x)$ as the product of three algebraic factors, and hence show that the equation $f(x)=0$ has only two real roots;

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1. find the set of values of $x$ for which $f(x)>0$.

Suggestion

You could use this applet to try some values. The green curve is $y=f(x)$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/pyUm9gmV/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Could we find two equations involving $a$ and $b$ that must be simultaneously true?

How would the Factor Theorem help us here?

How can we find the number of real roots of a quadratic equation?

Could sketching the graph of $f(x)$ help with this?

Solution

Substituting

$x=-\frac{1}{2}$

we have

$$\begin{align*} f\left(-\frac{1}{2}\right) &= 0\\ \implies\quad \frac{2}{16}-\frac{a}{8}+\frac{b}{4}+2-4 &= 0\\ \iff\quad 1 - a + 2b - 16 &= 0\\ \iff\quad -a+2b &= 15 \end{align*}$$

Also we have that

$$\begin{align*} f(2) &= 0\\ \implies\quad 32+8a+4b-8-4 &= 0\\ \iff\quad 2a+b &= -5 \end{align*}$$

We now have two simultaneous equations in $a$ and $b$. If we add twice the first to the second, to eliminate $a$, we see that $5b = 25$ and so $b = 5$.

We can then use either equation to find that $a = -5$.

Since $f\left(-\frac{1}{2}\right)=0$ and $f(2)=0$ we know that $2x+1$ and $x-2$ are two factors of $f(x)$.

So $f(x)=2x^4 - 5x^3 + 5x^2 - 4x - 4 =(2x+1)(x-2)g(x) = (2x^2 - 3x -2)g(x),$ where $g(x)$ is a quadratic polynomial. By inspection, we can see that $g(x)=x^2-x+2$.

Can you explain, ‘by inspection’?

To find the expression for $g(x)$, let’s write $g(x)=ax^2+bx+c$ and try to find $a$, $b$ and $c$ such that $(2x^2 - 3x -2)(ax^2+bx+c)=2x^4 - 5x^3 + 5x^2 - 4x - 4.$ We can either multiply out the brackets on the left hand side, group the terms and find what $a$, $b$ and $c$ must be, or we can try to `inspect’ the equation.

For example, the only terms that contribute to the $x^4$ term will be $2x^2$ from the first bracket and $ax^2$ from the second.

This tells us that $a=1$. Similarly, the constant term must come from the constant terms in both of the brackets, so we find that $-2c=-4$ and so $c=2$.

To find $b$, we check the coefficient of $x$, so that $-2b-3c=-4$. We know already that $c=2$, so we see that $b=-1$. So $g(x)=x^2-x+2$.

Therefore we have $f(x)=(2x+1)(x-2)(x^2-x+2).$

To show that $f(x)$ only has two real roots we need to show that $g(x)=x^2-x+2$ has no real roots.

The discriminant of $g(x)$ is $1 - 4 \times 2 = -7$, and since this is negative, we see that $g(x)$ has no real roots.

So, we can conclude that $f(x)$ has exactly two real roots, which are $x=-\dfrac{1}{2}$ and $x=2$.

If we sketch the graph of $f(x)$, we’ll clearly see the regions for which $f(x)>0$. We know that there are only two real roots, and we observe that the $y$-intercept is $-4$.

So the graph is

From this, we can see that the values of $x$ for which $f(x)>0$ are $x<-\dfrac{1}{2}$ and $x>2$.