In this collection we present 20 practice problems on higher-degree polynomial inequalities, arranged in order of increasing difficulty. Each exercise shows how to factor the polynomial, identify its roots, analyse the sign of each factor, and correctly determine the solution set.
The underlying principle is always the same: a polynomial inequality is solved by studying the sign of the polynomial. When the polynomial is written as a product of factors, the overall sign is obtained by combining the signs of the individual factors.
Exercise 1 — level ★☆☆☆☆
Solve the inequality:
\[ x^3-x>0 \]
Answer
\[ x\in(-1,0)\cup(1,+\infty) \]
Solution
We factor the polynomial by factoring out \(x\):
\[ x^3-x=x(x^2-1) \]
The difference of squares \(x^2-1\) factors as:
\[ x^2-1=(x-1)(x+1) \]
The inequality therefore becomes:
\[ x(x-1)(x+1)>0 \]
The roots are:
\[ x=-1,\qquad x=0,\qquad x=1 \]
These values partition the real line into the intervals:
\[ (-\infty,-1),\quad (-1,0),\quad (0,1),\quad (1,+\infty) \]
We analyse the sign of the product \(x(x-1)(x+1)\):
| Interval | Sign of the product |
|---|---|
| \((-\infty,-1)\) | \(-\) |
| \((-1,0)\) | \(+\) |
| \((0,1)\) | \(-\) |
| \((1,+\infty)\) | \(+\) |
The inequality requires the product to be positive, i.e.:
\[ x(x-1)(x+1)>0 \]
We therefore select the intervals where the sign is positive. Since the inequality is strict, the roots are not included.
The solution is:
\[ x\in(-1,0)\cup(1,+\infty) \]
Exercise 2 — level ★☆☆☆☆
Solve the inequality:
\[ x^3-4x\leq 0 \]
Answer
\[ x\in(-\infty,-2]\cup[0,2] \]
Solution
We factor out \(x\):
\[ x^3-4x=x(x^2-4) \]
We factor the difference of squares:
\[ x^2-4=(x-2)(x+2) \]
This gives:
\[ x(x-2)(x+2)\leq 0 \]
The roots are:
\[ x=-2,\qquad x=0,\qquad x=2 \]
We analyse the sign of the product on the four intervals determined by the roots.
| Interval | Sign |
|---|---|
| \((-\infty,-2)\) | \(-\) |
| \((-2,0)\) | \(+\) |
| \((0,2)\) | \(-\) |
| \((2,+\infty)\) | \(+\) |
The inequality requires:
\[ x(x-2)(x+2)\leq 0 \]
We therefore select the intervals where the product is negative or zero.
Since the inequality uses the symbol \(\leq\), the roots are included.
The solution is:
\[ x\in(-\infty,-2]\cup[0,2] \]
Exercise 3 — level ★☆☆☆☆
Solve the inequality:
\[ x^3-9x\geq 0 \]
Answer
\[ x\in[-3,0]\cup[3,+\infty) \]
Solution
We factor out \(x\):
\[ x^3-9x=x(x^2-9) \]
We factor:
\[ x^2-9=(x-3)(x+3) \]
The inequality becomes:
\[ x(x-3)(x+3)\geq 0 \]
The roots are:
\[ -3,\qquad 0,\qquad 3 \]
Since all roots are simple, the sign changes at each one.
| Interval | Sign |
|---|---|
| \((-\infty,-3)\) | \(-\) |
| \((-3,0)\) | \(+\) |
| \((0,3)\) | \(-\) |
| \((3,+\infty)\) | \(+\) |
The inequality requires a positive or zero sign.
We therefore include the roots and select the positive intervals:
\[ x\in[-3,0]\cup[3,+\infty) \]
Exercise 4 — level ★★☆☆☆
Solve the inequality:
\[ x^3+2x^2-3x<0 \]
Answer
\[ x\in(-\infty,-3)\cup(0,1) \]
Solution
We factor out \(x\):
\[ x^3+2x^2-3x=x(x^2+2x-3) \]
We factor the trinomial:
\[ x^2+2x-3=(x+3)(x-1) \]
Therefore:
\[ x(x+3)(x-1)<0 \]
The roots are:
\[ x=-3,\qquad x=0,\qquad x=1 \]
We analyse the sign:
| Interval | Sign |
|---|---|
| \((-\infty,-3)\) | \(-\) |
| \((-3,0)\) | \(+\) |
| \((0,1)\) | \(-\) |
| \((1,+\infty)\) | \(+\) |
The inequality requires the product to be negative.
Since the symbol is \(<\), the roots are not included.
The solution is:
\[ x\in(-\infty,-3)\cup(0,1) \]
Exercise 5 — level ★★☆☆☆
Solve the inequality:
\[ x^4-5x^2+4\leq 0 \]
Answer
\[ x\in[-2,-1]\cup[1,2] \]
Solution
The inequality contains only even powers of \(x\). We substitute:
\[ t=x^2 \]
This gives:
\[ t^2-5t+4\leq 0 \]
We factor the trinomial:
\[ t^2-5t+4=(t-1)(t-4) \]
Therefore:
\[ (t-1)(t-4)\leq 0 \]
The product is less than or equal to zero when:
\[ 1\leq t\leq 4 \]
Since \(t=x^2\), we need to solve:
\[ 1\leq x^2\leq 4 \]
This compound inequality is equivalent to:
\[ |x|\geq 1 \qquad \text{and} \qquad |x|\leq 2 \]
Therefore:
\[ x\in[-2,-1]\cup[1,2] \]
Exercise 6 — level ★★☆☆☆
Solve the inequality:
\[ x^4-1>0 \]
Answer
\[ x\in(-\infty,-1)\cup(1,+\infty) \]
Solution
We factor the polynomial as a difference of squares:
\[ x^4-1=(x^2-1)(x^2+1) \]
Factoring further:
\[ x^2-1=(x-1)(x+1) \]
Hence:
\[ x^4-1=(x-1)(x+1)(x^2+1) \]
We observe that:
\[ x^2+1>0 \qquad \forall x\in\mathbb{R} \]
Therefore the sign of the product depends solely on:
\[ (x-1)(x+1) \]
The inequality is thus equivalent to:
\[ (x-1)(x+1)>0 \]
The product is positive outside the roots \(-1\) and \(1\):
\[ x<-1 \qquad \text{or} \qquad x>1 \]
Since the inequality is strict, the roots are not included.
The solution is:
\[ x\in(-\infty,-1)\cup(1,+\infty) \]
Exercise 7 — level ★★☆☆☆
Solve the inequality:
\[ (x-2)^2(x+1)<0 \]
Answer
\[ x\in(-\infty,-1) \]
Solution
The polynomial is already written as a product of factors:
\[ (x-2)^2(x+1)<0 \]
The roots are:
\[ x=2 \qquad \text{and} \qquad x=-1 \]
The root \(x=2\) has multiplicity \(2\), so it does not produce a sign change.
The factor \((x-2)^2\) is always non-negative and vanishes only at \(x=2\).
For \(x\neq 2\), it is positive. Therefore, away from \(x=2\), the sign of the product depends entirely on the factor:
\[ x+1 \]
We have:
\[ x+1<0 \iff x<-1 \]
Since the inequality is strict, the roots are not included.
The solution is:
\[ x\in(-\infty,-1) \]
Exercise 8 — level ★★☆☆☆
Solve the inequality:
\[ (x+2)^2(x-3)\geq 0 \]
Answer
\[ x\in\{-2\}\cup[3,+\infty) \]
Solution
The inequality is:
\[ (x+2)^2(x-3)\geq 0 \]
The roots are:
\[ x=-2 \qquad \text{and} \qquad x=3 \]
The factor \((x+2)^2\) is a perfect square, so:
\[ (x+2)^2\geq 0 \qquad \forall x\in\mathbb{R} \]
It vanishes only at \(x=-2\), and is strictly positive for every \(x\neq -2\).
For \(x\neq -2\), the sign of the product therefore depends on the factor:
\[ x-3 \]
We have:
\[ x-3\geq 0 \iff x\geq 3 \]
Furthermore, the product also equals zero at \(x=-2\). Since the inequality uses the symbol \(\geq\), this value must be included.
The solution is:
\[ x\in\{-2\}\cup[3,+\infty) \]
Exercise 9 — level ★★★☆☆
Solve the inequality:
\[ x^4-6x^2+8>0 \]
Answer
\[ x\in(-\infty,-2)\cup(-\sqrt{2},\sqrt{2})\cup(2,+\infty) \]
Solution
The inequality contains only even powers of \(x\). We substitute:
\[ t=x^2 \]
This gives:
\[ t^2-6t+8>0 \]
We factor the trinomial:
\[ t^2-6t+8=(t-2)(t-4) \]
Therefore:
\[ (t-2)(t-4)>0 \]
The product is positive outside the roots \(2\) and \(4\), hence:
\[ t<2 \qquad \text{or} \qquad t>4 \]
Since \(t=x^2\), we need to solve:
\[ x^2<2 \qquad \text{or} \qquad x^2>4 \]
The first inequality gives:
\[ -\sqrt{2}<x<\sqrt{2} \]
The second inequality gives:
\[ x<-2 \qquad \text{or} \qquad x>2 \]
Taking the union of both solution sets:
\[ x\in(-\infty,-2)\cup(-\sqrt{2},\sqrt{2})\cup(2,+\infty) \]
Exercise 10 — level ★★★☆☆
Solve the inequality:
\[ x^4-10x^2+9\geq 0 \]
Answer
\[ x\in(-\infty,-3]\cup[-1,1]\cup[3,+\infty) \]
Solution
We substitute:
\[ t=x^2 \]
The inequality becomes:
\[ t^2-10t+9\geq 0 \]
We factor the trinomial:
\[ t^2-10t+9=(t-1)(t-9) \]
Therefore:
\[ (t-1)(t-9)\geq 0 \]
The product is non-negative when:
\[ t\leq 1 \qquad \text{or} \qquad t\geq 9 \]
Since \(t=x^2\), we obtain:
\[ x^2\leq 1 \qquad \text{or} \qquad x^2\geq 9 \]
Solving each:
\[ x^2\leq 1 \iff -1\leq x\leq 1 \]
and:
\[ x^2\geq 9 \iff x\leq -3 \quad \text{or} \quad x\geq 3 \]
The solution is therefore:
\[ x\in(-\infty,-3]\cup[-1,1]\cup[3,+\infty) \]
Exercise 11 — level ★★★☆☆
Solve the inequality:
\[ x^3-3x^2-4x+12\geq 0 \]
Answer
\[ x\in[-2,2]\cup[3,+\infty) \]
Solution
We factor by grouping:
\[ x^3-3x^2-4x+12=x^2(x-3)-4(x-3) \]
We factor out the common factor \(x-3\):
\[ x^2(x-3)-4(x-3)=(x-3)(x^2-4) \]
We factor the difference of squares:
\[ x^2-4=(x-2)(x+2) \]
The inequality therefore becomes:
\[ (x-3)(x-2)(x+2)\geq 0 \]
The roots are:
\[ x=-2,\qquad x=2,\qquad x=3 \]
We analyse the sign of the product.
| Interval | Sign |
|---|---|
| \((-\infty,-2)\) | \(-\) |
| \((-2,2)\) | \(+\) |
| \((2,3)\) | \(-\) |
| \((3,+\infty)\) | \(+\) |
The inequality requires a positive or zero sign. Since the symbol is \(\geq\), the roots are included.
The solution is:
\[ x\in[-2,2]\cup[3,+\infty) \]
Exercise 12 — level ★★★☆☆
Solve the inequality:
\[ x^3+3x^2-4\leq 0 \]
Answer
\[ x\in(-\infty,1] \]
Solution
We look for an integer root of the polynomial:
\[ P(x)=x^3+3x^2-4 \]
Testing \(x=1\):
\[ P(1)=1+3-4=0 \]
So \(x=1\) is a root, and \(x-1\) is a factor of the polynomial.
Dividing \(x^3+3x^2+0x-4\) by \(x-1\), we obtain:
\[ x^3+3x^2-4=(x-1)(x^2+4x+4) \]
The trinomial is a perfect square:
\[ x^2+4x+4=(x+2)^2 \]
Therefore:
\[ x^3+3x^2-4=(x-1)(x+2)^2 \]
The inequality becomes:
\[ (x-1)(x+2)^2\leq 0 \]
The factor \((x+2)^2\) is always non-negative and vanishes only at \(x=-2\).
For \(x\neq -2\), the sign of the product depends on the factor \(x-1\).
We have:
\[ x-1\leq 0 \iff x\leq 1 \]
A closer look is needed: for \(x<1\), the factor \(x-1\) is negative while \((x+2)^2\) is positive, except at \(x=-2\) where the entire product equals zero.
At \(x=1\) the product is also zero.
The inequality is therefore satisfied for all \(x\leq 1\).
The solution is:
\[ x\in(-\infty,1] \]
Exercise 13 — level ★★★★☆
Solve the inequality:
\[ x^4-2x^3-x^2+2x\geq 0 \]
Answer
\[ x\in(-\infty,-1]\cup[0,1]\cup[2,+\infty) \]
Solution
We factor out \(x\):
\[ x^4-2x^3-x^2+2x=x(x^3-2x^2-x+2) \]
We factor the cubic by grouping:
\[ x^3-2x^2-x+2=x^2(x-2)-1(x-2) \]
Factoring out \(x-2\):
\[ x^3-2x^2-x+2=(x-2)(x^2-1) \]
Factoring the difference of squares:
\[ x^2-1=(x-1)(x+1) \]
Therefore:
\[ x^4-2x^3-x^2+2x=x(x-2)(x-1)(x+1) \]
The inequality becomes:
\[ x(x-2)(x-1)(x+1)\geq 0 \]
The roots are:
\[ -1,\qquad 0,\qquad 1,\qquad 2 \]
We analyse the sign of the product.
| Interval | Sign |
|---|---|
| \((-\infty,-1)\) | \(+\) |
| \((-1,0)\) | \(-\) |
| \((0,1)\) | \(+\) |
| \((1,2)\) | \(-\) |
| \((2,+\infty)\) | \(+\) |
Note: The signs can be checked using a test value; since all roots are simple, the sign changes at each root. For example, at \(x=3\) all factors are positive, confirming the sign in the last interval is positive. Since all roots are simple, the sign changes at each one.
The inequality requires:
\[ x(x-2)(x-1)(x+1)\geq 0 \]
We therefore select the intervals where the sign is positive or zero.
The solution is:
\[ x\in(-\infty,-1]\cup[0,1]\cup[2,+\infty) \]
Exercise 14 — level ★★★★☆
Solve the inequality:
\[ x^5-x^4-4x^3+4x^2<0 \]
Answer
\[ x\in(-\infty,-2)\cup(1,2) \]
Solution
We factor out \(x^2\):
\[ x^5-x^4-4x^3+4x^2=x^2(x^3-x^2-4x+4) \]
We factor the cubic by grouping:
\[ x^3-x^2-4x+4=x^2(x-1)-4(x-1) \]
Factoring out \(x-1\):
\[ x^3-x^2-4x+4=(x-1)(x^2-4) \]
We factor the difference of squares:
\[ x^2-4=(x-2)(x+2) \]
Therefore:
\[ x^5-x^4-4x^3+4x^2=x^2(x-1)(x-2)(x+2) \]
The inequality becomes:
\[ x^2(x-1)(x-2)(x+2)<0 \]
The roots are:
\[ -2,\qquad 0,\qquad 1,\qquad 2 \]
The root \(x=0\) has multiplicity \(2\), so the sign does not change when crossing \(0\).
The remaining roots are simple, so the sign changes at each of them.
We analyse the sign of the product. Since \(x=0\) has even multiplicity, the sign is unchanged there; crossing the simple roots \(-2\), \(1\), and \(2\) each produces a sign change.
| Interval | Sign |
|---|---|
| \((-\infty,-2)\) | \(-\) |
| \((-2,0)\) | \(+\) |
| \((0,1)\) | \(+\) |
| \((1,2)\) | \(-\) |
| \((2,+\infty)\) | \(+\) |
The inequality requires a strictly negative sign. Since the symbol is \(<\), the roots are not included.
The solution is:
\[ x\in(-\infty,-2)\cup(1,2) \]
Exercise 15 — level ★★★★☆
Solve the inequality:
\[ (x-1)^3(x+2)^2\leq 0 \]
Answer
\[ x\in(-\infty,1] \]
Solution
The inequality is already written as a product of factors:
\[ (x-1)^3(x+2)^2\leq 0 \]
The roots are:
\[ x=1,\qquad x=-2 \]
The root \(x=1\) has multiplicity \(3\), which is odd, so the sign changes when crossing it.
The root \(x=-2\) has multiplicity \(2\), which is even, so the sign does not change when crossing it.
We also note that:
\[ (x+2)^2\geq 0 \]
for every \(x\in\mathbb{R}\). For \(x\neq -2\) this factor is strictly positive. Therefore, away from \(x=-2\), the sign of the product depends on the factor:
\[ (x-1)^3 \]
Since an odd power preserves the sign of its base, we have:
\[ (x-1)^3<0 \iff x<1 \]
Furthermore, the product vanishes at \(x=-2\) and at \(x=1\). Since the inequality uses the symbol \(\leq\), the roots are included.
The solution is therefore:
\[ x\in(-\infty,1] \]
Exercise 16 — level ★★★★☆
Solve the inequality:
\[ x^6-7x^3+6>0 \]
Answer
\[ x\in(-\infty,1)\cup(\sqrt[3]{6},+\infty) \]
Solution
The polynomial involves the powers \(x^6\) and \(x^3\). We substitute:
\[ t=x^3 \]
Then:
\[ x^6=(x^3)^2=t^2 \]
The inequality becomes:
\[ t^2-7t+6>0 \]
We factor the trinomial:
\[ t^2-7t+6=(t-1)(t-6) \]
Therefore:
\[ (t-1)(t-6)>0 \]
The product is positive when both factors have the same sign, i.e.:
\[ t<1 \qquad \text{or} \qquad t>6 \]
Returning to the original variable \(x\):
\[ x^3<1 \qquad \text{or} \qquad x^3>6 \]
Since the function \(x\mapsto x^3\) is strictly increasing on \(\mathbb{R}\), we can take cube roots without reversing the inequalities:
\[ x<1 \qquad \text{or} \qquad x>\sqrt[3]{6} \]
Since the inequality is strict, the endpoints are not included.
The solution is:
\[ x\in(-\infty,1)\cup(\sqrt[3]{6},+\infty) \]
Exercise 17 — level ★★★★☆
Solve the inequality:
\[ x^4-16x^2<0 \]
Answer
\[ x\in(-4,0)\cup(0,4) \]
Solution
We factor out \(x^2\):
\[ x^4-16x^2=x^2(x^2-16) \]
We factor the difference of squares:
\[ x^2-16=(x-4)(x+4) \]
The inequality becomes:
\[ x^2(x-4)(x+4)<0 \]
The roots are:
\[ x=-4,\qquad x=0,\qquad x=4 \]
The root \(x=0\) has multiplicity \(2\), arising from the factor \(x^2\), so the sign does not change when crossing it.
Furthermore:
\[ x^2\geq 0 \]
for every \(x\in\mathbb{R}\), and it is strictly positive for \(x\neq 0\). Away from \(x=0\), the sign of the product therefore depends on:
\[ (x-4)(x+4) \]
The product \((x-4)(x+4)\) is negative between the two roots:
\[ -4<x<4 \]
However, we must exclude \(x=0\), since the full product equals zero there, whereas the inequality requires a strictly negative value.
The solution is:
\[ x\in(-4,0)\cup(0,4) \]
Exercise 18 — level ★★★★☆
Solve the inequality:
\[ (x^2-3x+2)(x^2+2x+5)\geq 0 \]
Answer
\[ x\in(-\infty,1]\cup[2,+\infty) \]
Solution
We study the two factors separately:
\[ x^2-3x+2 \qquad \text{and} \qquad x^2+2x+5 \]
The first trinomial factors readily:
\[ x^2-3x+2=(x-1)(x-2) \]
Now consider the second trinomial:
\[ x^2+2x+5 \]
We compute its discriminant:
\[ \Delta=2^2-4\cdot1\cdot5=4-20=-16 \]
The discriminant is negative and the leading coefficient is positive. Consequently:
\[ x^2+2x+5>0 \qquad \forall x\in\mathbb{R} \]
The inequality is therefore equivalent to:
\[ (x-1)(x-2)\geq 0 \]
The product of two linear factors is non-negative outside the roots:
\[ x\leq 1 \qquad \text{or} \qquad x\geq 2 \]
Since the symbol is \(\geq\), the roots are included.
The solution is:
\[ x\in(-\infty,1]\cup[2,+\infty) \]
Exercise 19 — level ★★★★★
Solve the inequality:
\[ x^5-5x^4+6x^3\leq 0 \]
Answer
\[ x\in(-\infty,0]\cup[2,3] \]
Solution
We factor out \(x^3\):
\[ x^5-5x^4+6x^3=x^3(x^2-5x+6) \]
We factor the trinomial:
\[ x^2-5x+6=(x-2)(x-3) \]
Therefore:
\[ x^5-5x^4+6x^3=x^3(x-2)(x-3) \]
The inequality becomes:
\[ x^3(x-2)(x-3)\leq 0 \]
The roots are:
\[ x=0,\qquad x=2,\qquad x=3 \]
The root \(x=0\) has multiplicity \(3\), which is odd, so the sign changes when crossing it.
The roots \(x=2\) and \(x=3\) are also simple, so the sign changes at each of them as well.
We analyse the sign on each interval:
| Interval | Sign |
|---|---|
| \((-\infty,0)\) | \(-\) |
| \((0,2)\) | \(+\) |
| \((2,3)\) | \(-\) |
| \((3,+\infty)\) | \(+\) |
The inequality requires:
\[ x^3(x-2)(x-3)\leq 0 \]
We therefore select the intervals where the product is negative or zero.
Since the symbol is \(\leq\), the roots are included.
The solution is:
\[ x\in(-\infty,0]\cup[2,3] \]
Exercise 20 — level ★★★★★
Solve the inequality:
\[ x^4-3x^3-4x^2+12x\leq 0 \]
Answer
\[ x\in[-2,0]\cup[2,3] \]
Solution
We factor out \(x\):
\[ x^4-3x^3-4x^2+12x=x(x^3-3x^2-4x+12) \]
We factor the cubic by grouping:
\[ x^3-3x^2-4x+12=x^2(x-3)-4(x-3) \]
We factor out the common factor \(x-3\):
\[ x^2(x-3)-4(x-3)=(x-3)(x^2-4) \]
We factor the difference of squares:
\[ x^2-4=(x-2)(x+2) \]
Therefore:
\[ x^4-3x^3-4x^2+12x=x(x-3)(x-2)(x+2) \]
The inequality becomes:
\[ x(x-3)(x-2)(x+2)\leq 0 \]
The roots are:
\[ x=-2,\qquad x=0,\qquad x=2,\qquad x=3 \]
All roots are simple, so the sign changes at each one.
We analyse the sign of the product:
| Interval | Sign |
|---|---|
| \((-\infty,-2)\) | \(+\) |
| \((-2,0)\) | \(-\) |
| \((0,2)\) | \(+\) |
| \((2,3)\) | \(-\) |
| \((3,+\infty)\) | \(+\) |
The inequality requires a negative or zero sign.
Since the symbol is \(\leq\), the roots are included.
The solution is:
\[ x\in[-2,0]\cup[2,3] \]