4.4: Solve Polynomial Equations by Factoring (2024)

general guidelines for factoring polynomials

Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Step 2: Determine the number of terms in the polynomial.

1. Factor four-term polynomials by grouping.
2. Factor trinomials (3 terms) using “trial and error” or the AC method.
3. Factor binomials (2 terms) using the following special products:
   • Difference of squares:\(a^{2}−b^{2}=(a+b)(a−b)\)
   • Sum of squares: \(a^{2}+b^{2}\)  no general formula
   • Difference of cubes: \(a^{3}−b^{3}=(a−b)(a^{2}+ab+b^{2})\)
   • Sum of cubes: \(a^{3}+b^{3}=(a+b)(a^{2}−ab+b^{2})\)

Step 3: Look for factors that can be factored further.

Step 4: Check by multiplying.

Example \(\PageIndex{1}\):

Factor \(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x\).

Solution

This four-term polynomial has a GCF of \(2x\). Factor this out first.

\(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x = 2 x \left( 27 x ^ { 3 } - 18 x ^ { 2 } - 12 x + 8 \right)\)

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

Answer

\(2 x ( 3 x - 2 ) ^ { 2 } ( 3 x + 2 )\). The check is left to the reader.

Example \(\PageIndex{2}\):

Factor: \(x ^ { 4 } - 3 x ^ { 2 } - 4\).

Solution

This trinomial does not have a GCF.

\(\begin{aligned} x ^ { 4 } - 3 x ^ { 2 } - 4 & = \left( x ^ { 2 } \quad\right) \left( x ^ { 2 }\quad \right) \\ & = \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } - 4 \right) \quad\color{Cerulean} { Difference\: of\: squares } \\ & = \left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 ) \end{aligned}\)

The factor \(\left( x ^ { 2 } + 1 \right)\) is prime and the trinomial is completlely factored.

Answer:

\(\left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 )\)

Example \(\PageIndex{3}\):

Factor: \(x ^ { 6 } + 6 x ^ { 3 } - 16\).

Solution

Begin by factoring \(x ^ { 6 } = x ^ { 3 } \cdot x ^ { 3 }\) and look for the factors of \(16\) that add to \(6\).

\(\begin{aligned} x ^ { 6 } + 6 x ^ { 3 } - 16 & = \left( x ^ { 3 }\quad \right) \left( x ^ { 3 }\quad \right) \\ & = \left( x ^ { 3 } - 2 \right) \left( x ^ { 3 } + 8 \right)\quad \color{Cerulean} { sum\:of\:cubes } \\ & = \left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } - 2 x + 4 \right) \end{aligned}\)

The factor \(\left( x ^ { 3 } - 2 \right)\) cannot be factored any further using integers and the factorization is complete.

Answer:

\(\left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

Exercise \(\PageIndex{1}\)

Factor: \(9 x ^ { 4 } + 17 x ^ { 2 } - 2\)

Answer

\(( 3 x + 1 ) ( 3 x - 1 ) \left( x ^ { 2 } + 2 \right)\)

www.youtube.com/v/lsAP8_UgUx0

Example \(\PageIndex{4}\)

Solve: \(2 x ( x - 4 ) ( 5 x + 3 ) = 0\).

Solution

Set each variable factor equal to zero and solve.

\(\begin{aligned} 2 x & = 0 \\ \frac { 2 x } { \color{Cerulean}{2} } & = \frac { 0 } {\color{Cerulean}{ 2} } \\ x & = 0 \end{aligned}\) or \(\begin{aligned} x - 4 & = 0 \\ x & = 4 \end{aligned}\) or \(\begin{aligned} 5 x + 3 & = 0 \\ \frac { 5 x } { \color{Cerulean}{5} } & = \frac { - 3 } {\color{Cerulean}{ 5} } \\ x & = - \frac { 3 } { 5 } \end{aligned}\)

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

Answer:

The solutions are \(0, 4\), and \(\frac{−3}{5}\).

Example \(\PageIndex{5}\)

Solve: \(4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0\).

Solution

Begin by factoring the left side completely.

\(\begin{aligned} 4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 & = 0 \quad\color{Cerulean}{Factor\:by\:grouping.} \\ x ^ { 2 } ( 4 x - 1 ) - 25 ( 4 x - 1 ) & = 0 \\ ( 4 x - 1 ) \left( x ^ { 2 } - 25 \right) & = 0\quad\color{Cerulean}{Factor\:as\:a\:difference\:of\:squares.} \\ ( 4 x - 1 ) ( x + 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Set each factor equal to zero and solve.

\(\begin{array} { r } { 4 x - 1 = 0 } \\ { 4 x = 1 } \\ { x = \frac { 1 } { 4 } } \end{array}\) or \(\begin{aligned} x + 5 & = 0 \\ x & = - 5 \end{aligned}\) or \(\begin{aligned} x - 5 & = 0 \\ x & = 5 \end{aligned}\)

Answer:

The solutions are \(\frac{1}{4}, −5\), and \(5\).

Example \(\PageIndex{6}\)

Solve: \(15 x ^ { 2 } + 3 x - 8 = 5 x - 7\).

Solution

Step 1: Express the equation in standard form, equal to zero. In this example, subtract \(5x\) from and add \(7\) to both sides.

\(\begin{array} { l } { 15 x ^ { 2 } + 3 x - 8 = 5 x - 7 } \\ { 15 x ^ { 2 } - 2 x - 1 = 0 } \end{array}\)

Step 2: Factor the expression.

\((3x−1)(5x+1)=0\)

Step 3: Apply the zero-product property and set each variable factor equal to zero.

\(3x−1=0\)        or        \(5x+1=0\)

Step 4: Solve the resulting linear equations.

\(\begin{array} { r } { 3 x - 1 = 0 } \\ { 3 x = 1 } \\ { x = \frac { 1 } { 3 } } \end{array}\) or \(\begin{aligned} 5 x + 1 & = 0 \\ 5 x & = - 1 \\ x & = - \frac { 1 } { 5 } \end{aligned}\)

Answer:

The solutions are \(\frac{1}{3}\) and \(\frac{−1}{5}\). The check is optional.

Example \(\PageIndex{7}\):

Solve: \((3x+2)(x+1)=4\).

Solution

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(4\). However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

\(\begin{array} { r } { ( 3 x + 2 ) ( x + 1 ) = 4 } \\ { 3 x ^ { 2 } + 3 x + 2 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x - 2 = 0 } \end{array}\)

Once it is in standard form, we can factor and then set each factor equal to zero.

        \(\begin{array} { c } { ( 3 x - 1 ) ( x + 2 ) = 0 } \\ { 3 x - 1 = 0 \quad \text { or } \quad x + 2 = 0 } \\ \quad\quad{ 3 x = 1 }\quad\quad\quad\:\:\quad x=-2\\ { x = \frac { 1 } { 3 } }\quad\quad\quad\quad\quad\: \end{array}\)

Answer:

The solutions are \(\frac{1}{3}\) and \(−2\).

Example \(\PageIndex{8}\)

Find the roots: \(f ( x ) = ( x + 2 ) ^ { 2 } - 4\).

Solution

To find roots we set the function equal to zero and solve.

\(\begin{aligned} f ( x ) & = 0 \\ ( x + 2 ) ^ { 2 } - 4 & = 0 \\ x ^ { 2 } + 4 x + 4 - 4 & = 0 \\ x ^ { 2 } + 4 x & = 0 \\ x ( x + 4 ) & = 0 \end{aligned}\)

Next, set each factor equal to zero and solve.

\(\begin{aligned} x = 0 \quad \text { or } \quad x + 4 = 0 \\ x = - 4 \end{aligned}\)

We can show that these \(x\)-values are roots by evaluating.

\(\begin{aligned} f ( 0 ) & = ( 0 + 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)   \(\begin{aligned} f ( - 4 ) & = ( - 4 + 2 ) ^ { 2 } - 4 \\ & = ( - 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)

Answer:

The roots are \(0\) and \(−4\).

Example \(\PageIndex{9}\)

Find the roots: \(f ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4\).

Solution

To find roots we set the function equal to zero and solve.

\(\begin{aligned} f ( x ) & = 0 \\ x ^ { 4 } - 5 x ^ { 2 } + 4 & = 0 \\ \left( x ^ { 2 } - 1 \right) \left( x ^ { 2 } - 4 \right) & = 0 \\ ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \end{aligned}\)

Next, set each factor equal to zero and solve.

\(\begin{aligned} x + 1 & = 0 \\ x & = - 1 \end{aligned}\) or \(\begin{array} { r } { x - 1 = 0 } \\ { x = 1 } \end{array}\) or \(\begin{aligned} x + 2 & = 0 \\ x & = - 2 \end{aligned}\) or \(\begin{aligned} x - 2 & = 0 \\ x & = 2 \end{aligned}\)

Answer:

The roots are \(−1, 1, −2\), and \(2\).

Example \(\PageIndex{10}\)

Find the roots: \(f ( x ) = - x ^ { 2 } + 10 x - 25\).

Solution

To find roots we set the function equal to zero and solve.

\(\begin{aligned} f ( x ) & = 0 \\ - x ^ { 2 } + 10 x - 25 & = 0 \\ - \left( x ^ { 2 } - 10 x + 25 \right) & = 0 \\ - ( x - 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Next, set each variable factor equal to zero and solve.

\(\begin{aligned} x - 5 & = 0 \quad\quad\text { or }& x - 5 = 0 \\ & = 5 \quad &x= 5 \end{aligned}\)

A solution that is repeated twice is called a double root24. In this case, there is only one solution.

Answer:

The root is \(5\).

Exercise \(\PageIndex{2}\)

Find the roots of \(f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3\).

Answer

\(±1, −3\)

www.youtube.com/v/t1ShqqhoaVE

Example \(\PageIndex{11}\)

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\), where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(40\) feet.

Solution

We are asked to find the speed \(x\) where the safe stopping distance \(d(x)=40\) feet.

\(\begin{array} { c } { d ( x ) = 40 } \\ { \frac { 1 } { 20 } x ^ { 2 } + x = 40 } \end{array}\)

To solve for \(x\), rewrite the resulting equation in standard form. In this case, we will first multiply both sides by \(20\) to clear the fraction.

\(\begin{aligned} \color{Cerulean}{20}\color{black}{ \left( \frac { 1 } { 20 } x ^ { 2 } + x \right)} & = \color{Cerulean}{20}\color{black}{ (} 40 ) \\ x ^ { 2 } + 20 x & = 800 \\ x ^ { 2 } + 20 x - 800 & = 0 \end{aligned}\)

Next factor and then set each factor equal to zero.

\(\begin{aligned} x ^ { 2 } + 20 x - 800 & = 0 \\ ( x + 40 ) ( x - 20 ) & = 0 \\ x + 40 & = 0\quad or\quad \:x-20=0 \\ x & = - 40\quad \quad\quad\quad x=20 \end{aligned}\)

The negative answer does not make sense in the context of this problem. Consider \(x=20\) miles per hour to be the only solution.

Answer:

\(20\) miles per hour

Example \(\PageIndex{12}\)

Find a quadratic equation with integer coefficients, given solutions \(\frac{−3}{2}\) and \(\frac{1}{3}\).

Solution

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

\(\begin{aligned} x & = - \frac { 3 } { 2 } \\ 2 x & = - 3 \\ 2 x + 3 & = 0 \end{aligned}\) or \(\begin{aligned} x & = \frac { 1 } { 3 } \\ 3 x & = 1 \\ 3 x - 1 & = 0 \end{aligned}\)

The product of these linear factors is equal to zero when \(x=\frac{−3}{2}\) or \(x=\frac{1}{3}\).

\((2x+3)(3x−1)=0\)

Multiply the binomials and present the equation in standard form.

\(\begin{array} { r } { 6 x ^ { 2 } - 2 x + 9 x - 3 = 0 } \\ { 6 x ^ { 2 } + 7 x - 3 = 0 } \end{array}\)

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

Answer:

\(6 x ^ { 2 } + 7 x - 3 = 0\)

Example \(\PageIndex{13}\)

Find a polynomial function with real roots \(1, −2\), and \(2\).

Solution

Given solutions to \(f(x)=0\) we can find linear factors.

\(\begin{array} { r } { x = 1 } \\ { x - 1 = 0 } \end{array}\) or \(\begin{aligned} x & = - 2 \\ x + 2 & = 0 \end{aligned}\) or \(\begin{aligned} x & = 2 \\ x - 2 & = 0 \end{aligned}\)

Apply the zero-product property and multiply.

\(\begin{aligned} ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \\ ( x - 1 ) \left( x ^ { 2 } - 4 \right) & = 0 \\ x ^ { 3 } - 4 x - x ^ { 2 } + 4 & = 0 \\ x ^ { 3 } - x ^ { 2 } - 4 x + 4 & = 0 \end{aligned}\)

Answer:

\(f ( x ) = x ^ { 3 } - x ^ { 2 } - 4 x + 4\)

Exercise \(\PageIndex{3}\)

Find a polynomial equation with integer coefficients, given solutions \(\frac{1}{2}\) and \(\frac{−3}{4}\).

Answer

\(8 x ^ { 2 } + 2 x - 3 = 0\)

www.youtube.com/v/o4cuUWWEGdU

Exercise \(\PageIndex{4}\)

Factor completely.

1. \(50 x ^ { 2 } - 18\)
2. \(12 x ^ { 3 } - 3 x\)
3. \(10 x ^ { 3 } + 65 x ^ { 2 } - 35 x\)
4. \(15 x ^ { 4 } + 7 x ^ { 3 } - 4 x ^ { 2 }\)
5. \(6 a ^ { 4 } b - 15 a ^ { 3 } b ^ { 2 } - 9 a ^ { 2 } b ^ { 3 }\)
6. \(8 a ^ { 3 } b - 44 a ^ { 2 } b ^ { 2 } + 20 a b ^ { 3 }\)
7. \(36 x ^ { 4 } - 72 x ^ { 3 } - 4 x ^ { 2 } + 8 x\)
8. \(20 x ^ { 4 } + 60 x ^ { 3 } - 5 x ^ { 2 } - 15 x\)
9. \(3 x ^ { 5 } + 2 x ^ { 4 } - 12 x ^ { 3 } - 8 x ^ { 2 }\)
10. \(10 x ^ { 5 } - 4 x ^ { 4 } - 90 x ^ { 3 } + 36 x ^ { 2 }\)
11. \(x ^ { 4 } - 23 x ^ { 2 } - 50\)
12. \(2 x ^ { 4 } - 31 x ^ { 2 } - 16\)
13. \(- 2 x ^ { 5 } - 6 x ^ { 3 } + 8 x\)
14. \(- 36 x ^ { 5 } + 69 x ^ { 3 } + 27 x\)
15. \(54 x ^ { 5 } - 78 x ^ { 3 } + 24 x\)
16. \(4 x ^ { 6 } - 65 x ^ { 4 } + 16 x ^ { 2 }\)
17. \(x ^ { 6 } - 7 x ^ { 3 } - 8\)
18. \(x ^ { 6 } - 25 x ^ { 3 } - 54\)
19. \(3 x ^ { 6 } + 4 x ^ { 3 } + 1\)
20. \(27 x ^ { 6 } - 28 x ^ { 3 } + 1\)

Answer

1. \(2 ( 5 x + 3 ) ( 5 x - 3 )\)

3. \(5 x ( x + 7 ) ( 2 x - 1 )\)

5. \(3 a ^ { 2 } b ( 2 a + b ) ( a - 3 b )\)

7. \(4 x ( x - 2 ) ( 3 x + 1 ) ( 3 x - 1 )\)

9. \(x ^ { 2 } ( 3 x + 2 ) ( x + 2 ) ( x - 2 )\)

11. \(\left( x ^ { 2 } + 2 \right) ( x + 5 ) ( x - 5 )\)

13. \(- 2 x \left( x ^ { 2 } + 4 \right) ( x - 1 ) ( x + 1 )\)

15. \(6 x ( x + 1 ) ( x - 1 ) ( 3 x + 2 ) ( 3 x - 2 )\)

17. \(( x + 1 ) \left( x ^ { 2 } - x + 1 \right) ( x - 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

19. \(\left( 3 x ^ { 3 } + 1 \right) ( x + 1 ) \left( x ^ { 2 } - x + 1 \right)\)

Exercise \(\PageIndex{5}\)

Solve.

1. \(( 6 x - 5 ) ( x + 7 ) = 0\)
2. \(( x + 9 ) ( 3 x - 8 ) = 0\)
3. \(5 x ( 2 x - 5 ) ( 3 x + 1 ) = 0\)
4. \(4 x ( 5 x - 1 ) ( 2 x + 3 ) = 0\)
5. \(( x - 1 ) ( 2 x + 1 ) ( 3 x - 5 ) = 0\)
6. \(( x + 6 ) ( 5 x - 2 ) ( 2 x + 9 ) = 0\)
7. \(( x + 4 ) ( x - 2 ) = 16\)
8. \(( x + 1 ) ( x - 7 ) = 9\)
9. \(( 6 x + 1 ) ( x + 1 ) = 6\)
10. \(( 2 x - 1 ) ( x - 4 ) = 39\)
11. \(x ^ { 2 } - 15 x + 50 = 0\)
12. \(x ^ { 2 } + 10 x - 24 = 0\)
13. \(3 x ^ { 2 } + 2 x - 5 = 0\)
14. \(2 x ^ { 2 } + 9 x + 7 = 0\)
15. \(\frac { 1 } { 10 } x ^ { 2 } - \frac { 7 } { 15 } x - \frac { 1 } { 6 } = 0\)
16. \(\frac { 1 } { 4 } - \frac { 4 } { 9 } x ^ { 2 } = 0\)
17. \(6 x ^ { 2 } - 5 x - 2 = 30 x + 4\)
18. \(6 x ^ { 2 } - 9 x + 15 = 20 x - 13\)
19. \(5 x ^ { 2 } - 23 x + 12 = 4 ( 5 x - 3 )\)
20. \(4 x ^ { 2 } + 5 x - 5 = 15 ( 3 - 2 x )\)
21. \(( x + 6 ) ( x - 10 ) = 4 ( x - 18 )\)
22. \(( x + 4 ) ( x - 6 ) = 2 ( x + 4 )\)
23. \(4 x ^ { 3 } - 14 x ^ { 2 } - 30 x = 0\)
24. \(9 x ^ { 3 } + 48 x ^ { 2 } - 36 x = 0\)
25. \(\frac { 1 } { 3 } x ^ { 3 } - \frac { 3 } { 4 } x = 0\)
26. \(\frac { 1 } { 2 } x ^ { 3 } - \frac { 1 } { 50 } x = 0\)
27. \(- 10 x ^ { 3 } - 28 x ^ { 2 } + 48 x = 0\)
28. \(- 2 x ^ { 3 } + 15 x ^ { 2 } + 50 x = 0\)
29. \(2 x ^ { 3 } - x ^ { 2 } - 72 x + 36 = 0\)
30. \(4 x ^ { 3 } - 32 x ^ { 2 } - 9 x + 72 = 0\)
31. \(45 x ^ { 3 } - 9 x ^ { 2 } - 5 x + 1 = 0\)
32. \(x ^ { 3 } - 3 x ^ { 2 } - x + 3 = 0\)
33. \(x ^ { 4 } - 5 x ^ { 2 } + 4 = 0\)
34. \(4 x ^ { 4 } - 37 x ^ { 2 } + 9 = 0\)

Answer

1. \(- 7 , \frac { 5 } { 6 }\)

3. \(0 , \frac { 5 } { 2 } , - \frac { 1 } { 3 }\)

5. \(- \frac { 1 } { 2 } , 1 , \frac { 5 } { 3 }\)

7. \(- 6,4\)

9. \(- \frac { 5 } { 3 } , \frac { 1 } { 2 }\)

11. \(5,10\)

13. \(- \frac { 5 } { 3 } , 1\)

15. \(- \frac { 1 } { 3 } , 5\)

17. \(- \frac { 1 } { 6 } , 6\)

19. \(\frac { 3 } { 5 } , 8\)

21. \(2,6\)

23. \(0 , - \frac { 3 } { 2 } , 5\)

25. \(0 , \pm \frac { 3 } { 2 }\)

27. \(- 4,0 , \frac { 6 } { 5 }\)

29. \(\pm 6 , \frac { 1 } { 2 }\)

31. \(\pm \frac { 1 } { 3 } , \frac { 1 } { 5 }\)

33. \(\pm 1 , \pm 2\)

Exercise \(\PageIndex{6}\)

Find the roots of the given functions.

1. \(f ( x ) = x ^ { 2 } + 10 x - 24\)
2. \(f ( x ) = x ^ { 2 } - 14 x + 48\)
3. \(f ( x ) = - 2 x ^ { 2 } + 7 x + 4\)
4. \(f ( x ) = - 3 x ^ { 2 } + 14 x + 5\)
5. \(f ( x ) = 16 x ^ { 2 } - 40 x + 25\)
6. \(f ( x ) = 9 x ^ { 2 } - 12 x + 4\)
7. \(g ( x ) = 8 x ^ { 2 } + 3 x\)
8. \(g ( x ) = 5 x ^ { 2 } - 30 x\)
9. \(p ( x ) = 64 x ^ { 2 } - 1\)
10. \(q ( x ) = 4 x ^ { 2 } - 121\)
11. \(f ( x ) = \frac { 1 } { 5 } x ^ { 3 } - 1 x ^ { 2 } - \frac { 1 } { 20 } x + \frac { 1 } { 4 }\)
12. \(f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } - \frac { 4 } { 3 } x - 2\)
13. \(g ( x ) = x ^ { 4 } - 13 x ^ { 2 } + 36\)
14. \(g ( x ) = 4 x ^ { 4 } - 13 x ^ { 2 } + 9\)
15. \(f ( x ) = ( x + 5 ) ^ { 2 } - 1\)
16. \(g ( x ) = - ( x + 5 ) ^ { 2 } + 9\)
17. \(f ( x ) = - ( 3 x - 5 ) ^ { 2 }\)
18. \(g ( x ) = - ( x + 2 ) ^ { 2 } + 4\)

Answer

1. \(2 , - 12\)

3. \(- \frac { 1 } { 2 } , 4\)

5. \(\frac { 5 } { 4 }\)

7. \(- \frac { 3 } { 8 } , 0\)

9. \(\pm \frac { 1 } { 8 }\)

11. \(\pm \frac { 1 } { 2 } , 5\)

13. \(\pm 2 , \pm 3\)

15. \(- 6 , - 4\)

17. \(\frac { 5 } { 3 }\)

Exercise \(\PageIndex{7}\)

Given the graph of a function, determine the real roots.

1.

2.

3.

4.

5. The sides of a square measure \(x − 2\) units. If the area is \(36\) square units, then find \(x\).

6. The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

7. The profit in dollars generated by producing and selling n bicycles per week is given by the formula \(P ( n ) = - 5 n ^ { 2 } + 400 n - 6000\). How many bicycles must be produced and sold to break even?

8. The height in feet of an object dropped from the top of a \(64\)-foot building is given by \(h ( t ) = - 16 t ^ { 2 } + 64\) where \(t\) represents the time in seconds after it is dropped. How long will it take to hit the ground?

9. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

What is the length of each side of the cardboard sheet if the volume of the box is to be \(98\) cubic inches?

10. The height of a triangle is \(4\) centimeters less than twice the length of its base. If the total area of the triangle is \(48\) square centimeters, then find the lengths of the base and height.

11. A uniform border is to be placed around an \(8 × 10\) inch picture.

If the total area including the border must be \(168\) square inches, then how wide should the border be?

12. The area of a picture frame including a \(3\)-inch wide border is \(120\) square inches.

If the width of the inner area is \(2\) inches less than its length, then find the dimensions of the inner area.

13. Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\) where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(75\) feet.

14. A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula \(R ( n ) = 12 n - 0.6 n ^ { 2 }\) where \(n\) represents the number of palettes of product sold \((0 ≤ n < 20)\). Determine the number of palettes sold in a day if the revenue was \(45\) thousand dollars.

Answer

1. \(- 3 , - 1,0,2\)

3. \(- 2,3\)

5. \(8\) units

7. \(20\) or \(60\) bicycles

9. \(11\) in

11. \(2\) inches

13. \(30\) miles per hour

Exercise \(\PageIndex{8}\)

Find a polynomial equation with the given solutions.

1. \(-3,5\)
2. \(-1,8\)
3. \(2, \frac{1}{3}\)
4. \(- \frac { 3 } { 4 } , 5\)
5. \(0,-4\)
6. \(0,7\)
7. \(\pm 7\)
8. \(\pm 2\)
9. \(- 3,1,3\)
10. \(- 5 , - 1,1\)

Answer

1. \(x ^ { 2 } - 2 x - 15 = 0\)

3. \(3 x ^ { 2 } - 7 x + 2 = 0\)

5. \(x ^ { 2 } + 4 x = 0\)

7. \(x ^ { 2 } - 49 = 0\)

9. \(x ^ { 3 } - x ^ { 2 } - 9 x + 9 = 0\)

Exercise \(\PageIndex{9}\)

Find a function with the given roots.

1. \(\frac { 1 } { 2 } , \frac { 2 } { 3 }\)
2. \(\frac { 2 } { 5 } , - \frac { 1 } { 3 }\)
3. \(\pm \frac { 3 } { 4 }\)
4. \(\pm \frac { 5 } { 2 }\)
5. \(5\) double root
6. \(-3\) double root
7. \(-1,0,3\)
8. \(-5,0,2\)

Answer

1. \(f ( x ) = 6 x ^ { 2 } - 7 x + 2\)

3. \(f ( x ) = 16 x ^ { 2 } - 9\)

5. \(f ( x ) = x ^ { 2 } - 10 x + 25\)

7. \(f ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 3 x\)

Exercise \(\PageIndex{10}\)

Recall that if \(| X | = p\), then \(X=-p\) or \(X=p\). Use this to solve the following absolute value equations.

1. \(\left| x ^ { 2 } - 8 \right| = 8\)
2. \(\left| 2 x ^ { 2 } - 9 \right| = 9\)
3. \(\left| x ^ { 2 } - 2 x - 1 \right| = 2\)
4. \(\left| x ^ { 2 } - 8 x + 14 \right| = 2\)
5. \(\left| 2 x ^ { 2 } - 4 x - 7 \right| = 9\)
6. \(\left| x ^ { 2 } - 3 x - 9 \right| = 9\)

Answer

1. \(\pm 4,0\)

3. \(\pm 1,3\)

5. \(- 2,1,4\)

Exercise \(\PageIndex{11}\)

1. Explain to a beginning algebra student the difference between an equation and an expression.
2. What is the difference between a root and an \(x\)-intercept? Explain.
3. Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.
4. Research and discuss the fundamental theorem of algebra.

Answer

1. Answer may vary

3. Answer may vary