Inequalities with polynomials and rational expressionsAQA A-Level Further Maths: Revision notes

To solve $f(x)>0$ (or $<$, $\ge$, $\le$), where $f$ is a polynomial: move everything to one side, factorise, find the critical values (roots), then decide the sign on each interval with a sketch or sign table. Never divide by an expression containing $x$. For $p(x)=x^3-7x+6=(x-1)(x-2)(x+3)$ the positive leading coefficient means $p$ starts negative and changes sign at each root, so $p>0$ for $-3<x<1$ and $x>2$. For the quartic $q(x)=x^4-5x^2+4=(x^2-1)(x^2-4)$, put $u=x^2$ to factorise, then $q<0$ on $-2<x<-1$ and $1<x<2$. Use $\le$ or $\ge$ to include the roots, and $<$ or $>$ to exclude them.

A repeated root does not change the sign of the polynomial: the graph touches the axis there. For $(x-1)^2(x+1)>0$, the factor $(x-1)^2$ is non-negative, so the sign is decided by $x+1$: $x>-1$, but $x=1$ gives $0$ and must be excluded. For a non-strict inequality such as $(x-1)^2(x+1)\ge0$ the repeated root is included: $x\ge-1$, and $x=1$ is already within it. Example: $C_1:y=x^3-x$ lies above $C_2:y=x^2-1$ when $x^3-x^2-x+1=(x-1)^2(x+1)>0$, so $x>-1$, $x\ne1$; the curves touch at $(1,0)$ and cross at $(-1,0)$.

For $\left|\frac{ax+b}{cx+d}\right|<ex+f$, work algebraically:

• Domain and sign check: the denominator must not be $0$. Since the modulus is never negative, $|A|<B$ needs $B>0$.
• When $B>0$ you can square both sides (both sides are positive): $|A|<B\Leftrightarrow A^2<B^2$. Multiply through by the square of the denominator, which is positive, to clear fractions.
• Rearrange to $B^2-A^2>0$ and use the difference of two squares, $(B-A)(B+A)$, or move all terms to one side and factorise.
• Alternatively, $|A|<B\Leftrightarrow-B<A<B$, solving each part, when $A$ is a simple expression. For $|A|\ge B$: if $B\le0$ it is always true (where defined); if $B>0$, square.

Solve $\left|\frac{x+2}{x-1}\right|<x+1$.

1. The modulus is non-negative, so need $x+1>0$: $x>-1$. Also $x\ne1$.
2. Square and multiply by $(x-1)^2$: $(x+2)^2<(x+1)^2(x-1)^2=(x^2-1)^2$.
3. $(x^2-1)^2-(x+2)^2=(x^2-x-3)(x^2+x+1)>0$. Since $x^2+x+1=\left(x+\frac12\right)^2+\frac34>0$ always, $x^2-x-3>0$, so $x<\frac{1-\sqrt{13}}{2}$ or $x>\frac{1+\sqrt{13}}{2}$.
4. Combine with $x>-1$ (note $\frac{1-\sqrt{13}}{2}\approx-1.30$): the solution is $x>\frac{1+\sqrt{13}}{2}$. Check: $x=3$ gives $\left|\frac52\right|=2.5<4$, true; $x=2$ gives $4<3$, false.