College Algebra - Polynomial Inequalities

Give the solution set of the inequality

$x^{3} + 4x >7x^{2} + 28$

$\left { x | x > 7\right }$

$\left { x | x < 7\right }$

$\left { x | x< -2 \textrm{ or } x > 7\right }$

$\left { x | x< -2 \textrm{ or } 2 < x < 7\right }$

$\left { x | -2 < x< 2 \textrm{ or } x > 7\right }$

Explanation

To solve the polynomial inequality

$x^{3} + 4x >7x^{2} + 28$

it is necessary to get it into standard form by writing all terms at the same side. Subtract $7x^{2} + 28$ from both sides:

$x^{3} + 4x - (7x^{2} + 28 ) >7x^{2} + 28 - (7x^{2} + 28 )$

$x^{3}- 7x^{2} + 4x -28 >0$

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

$x^{3}- 7x^{2} + 4x -28 =0$

Factor the polynomial by grouping and taking out the GCFs:

$(x^{3}- 7x^{2} )+( 4x -28) =0$

$x^{2}(x - 7 ) + 4 (x - 7 )=0$

$(x^{2} + 4 )(x - 7 )=0$

The polynomial cannot be factored further, as $x^{2}+ 4$ , as the sum of squares, is prime. By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

$x^{2}+ 4 = 0$

$x^{2}+ 4 - 4 = 0 - 4$

$x^{2} = -4$ ,

which has no real solution.

Therefore, the only boundary of the intervals is 7, which divides the real numbers into two intervals, $(-\infty , 7)$ and $(7, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , 7)$ : Choose

$x^{3} + 4x >7x^{2} + 28$

$0 ^{3} + 4 (0) >7 \cdot 0^{2} + 28$

False - do not include $(-\infty , 7)$

$(7, \infty)$ : Choose

$x^{3} + 4x >7x^{2} + 28$

$8^{3} + 4 \cdot 8 >7 \cdot 8^{2} + 28$

$512+ 32 >7 \cdot 64 + 28$

True - include $(7, \infty)$

Also, since the inequality symbol does not allow for equality, 7 is not included in the solution. The solution set is $\left { x | x > 7\right }$ .

Which is correct graph of the following function?

$y\geq-3x^{2}$

Solutions

Wrong1

Wrong2

Wrong3

None of the other graphs.

Explanation

First graph $y=-3x^{2}$

Since the inequality is $\geq$ , this means that the line is included in the solution, as indicated by a solid line.

Shade the area outside of the parabola since it is a negative parabola, but greater than or equal to.

Solve

$5<x<\infty$

$-\infty<x<-2$

$5<x<\infty$

$-\infty<x<-2$

$-\infty<x<\infty$

$0<x<\infty$

Explanation

First we substract on each side

Now we can factor the left hand side

From this, we can see that the roots are at , and .

To make this inequality true, we will have two answers.

$5<x<\infty$

$-\infty<x<-2$

Give the solution set of the inequality

$(x- 5) (x+ 5 )^{2} \ge 0$

$\left { x | x = -5 \textrm{ or } x \ge 5 \right }$

$\left { x | x \ge 5 \right }$

$\left { x | x \le -5 \textrm{ or } -5 \le x \le 5 \right }$

$\left { x | -5 \le x \le 5 \right }$

$\left { x | x \le - 5 \right }$

Explanation

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- 5) (x+ 5 )^{2} \ge 0$

are the zeroes of the equation

$(x- 5) (x+ 5 )^{2} = 0$

By the Zero Product Property, either

in which case

and 5 are the boundaries of the intervals, which are $(-\infty , -5)$ , , and $(5, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -5)$ : Choose

$(x- 5) (x+ 5 )^{2} \ge 0$

$(-6- 5) (-6+ 5 )^{2} \ge 0$

$(-11) (-1 )^{2} \ge 0$

$(-11) (1 ) \ge 0$

$-11 \ge 0$

False - do not include $(-\infty , -5)$

: Choose

$(x- 5) (x+ 5 )^{2} \ge 0$

$(0- 5) (0+ 5 )^{2} \ge 0$

$( - 5) ( 5 )^{2} \ge 0$

$( - 5) (2 5 ) \ge 0$

$-125 \ge 0$

False - do not include

$(5, \infty)$ : Choose

$(x- 5) (x+ 5 )^{2} \ge 0$

$(6- 5) (6+ 5 )^{2} \ge 0$

$(1) (11 )^{2} \ge 0$

$(1) (121 ) \ge 0$

$121 \ge 0$

True - include $(5, \infty)$

Also, the boundary values themselves should be included, because the inequality symbol $\ge$ allows for equality.

The solution set is therefore $\left { x | x = -5 \textrm{ or } x \ge 5 \right }$ .

Give the solution set of the inequality

$(x- 4)^{2} (x+ 4 )>0$

$\left { x | -4 < x < 4 \textrm{ or }x> 4 \ \right }$

The set of all real numbers

$\left { x | x> -4 \ \right }$

$\left { x | -4 < x < 4 \ \right }$

$\left { x | x< -4\textrm{ or } -4 < x < 4 \ \right }$

Explanation

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- 4)^{2} (x+ 4 )>0$

are the zeroes of the equation

$(x- 4)^{2} (x+ 4 )= 0$

By the Zero Product Property, either

in which case

and 4 are the boundaries of the intervals, which are $(-\infty , -4)$ , , and $(4, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -4)$ : Choose

$(x- 4)^{2} (x+ 4 )> 0$

$(-5- 4)^{2} (-5+ 4 )> 0$

$(-9)^{2} (-1 )>0$

False - do not include $(-\infty , -4)$

: Choose

$(x- 4)^{2} (x+ 4 )> 0$

$(0- 4)^{2} (0+ 4 )>0$

$( - 4)^{2} (4) > 0$

True - include

$(4, \infty)$ : Choose

$(x- 4)^{2} (x+ 4 )> 0$

$(5- 4)^{2} (5+ 4 )>0$

$1^{2}(9)>0$

True - include $(4, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is therefore $\left { x | -4 < x < 4 \textrm{ or }x> 4 \ \right }$

Give the solution set of the inequality

$x^{3} + 7x^{2} < 4x+ 28$

$\left { x| x < -7 \textrm{ or } -2 < x < 2 \right }$

$\left { x| -7 < x < -2 \textrm{ or } x > 2 \right }$

$\left { x| x < -7 \right }$

$\left { x| x > -7 \right }$

$\left { x| -2 < x < 2 \right }$

Explanation

To solve the polynomial inequality

$x^{3} + 7x^{2} < 4x+ 28$

it is necessary to get it into standard form by writing all terms at the same side. Subtract from both sides:

$x^{3} + 7x^{2} - (4x+ 28) < 4x+ 28 - (4x+ 28)$

$x^{3} + 7x^{2} - 4x-28 < 0$

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

$x^{3} + 7x^{2} - 4x-28 = 0$

Factor the polynomial by grouping and taking out the GCFs:

$(x^{3} + 7x^{2}) - (4x+28) = 0$

$x^{2} (x + 7) - 4(x + 7) = 0$

$(x^{2} - 4)(x + 7) = 0$

$x^{2} - 4$ is the difference of squares and can be factored according to a pattern:

$(x^{2} - 2^{2})(x + 7) = 0$

By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

, , and 2 are the boundaries of the intervals, which are $(-\infty, -7)$ , , , and $(2, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty, -7)$ : Choose

$x^{3} + 7x^{2} < 4x+ 28$

$(-8)^{3} + 7 \cdot (-8)^{2} < 4(-8)+ 28$

$-512 + 7 \cdot 64 < -32+ 28$

True - include $(-\infty, -7)$

: Choose

$x^{3} + 7x^{2} < 4x+ 28$

$(-3)^{3} + 7 \cdot (-3)^{2} < 4(-3)+ 28$

$-27 + 7 \cdot 9 < -12 + 28$

False - do not include

: Choose

$x^{3} + 7x^{2} < 4x+ 28$

$0^{3} + 7 \cdot 0^{2} < 4 \cdot 0 + 28$

True - include

$(2, \infty)$ : Choose

$x^{3} + 7x^{2} < 4x+ 28$

$3^{3} + 7 \cdot 3^{2} < 4 \cdot 3+ 28$

$27 + 7 \cdot 9 < 12+ 28$

False - do not include $(2, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is $\left { x| x < -7 \textrm{ or } -2 < x < 2 \right }$

When is the inequality true?

this inequality cannot be satisfied.

Explanation

We can rewrite this as . In order to solve the inequality we need to factor the polynomial. Taking a couple educated guesses we find that this polynomial factors as . So now we have our inequality written as