Find the Roots/Zeros Using the Rational Roots Test 2x^3-7x^2-18x+63

$2 x^{3} - 7 x^{2} - 18 x + 63$

Step 1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 3, \pm 7, \pm 9, \pm 21, \pm 63$

$q = \pm 1, \pm 2$

Step 2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 7, \pm \frac{7}{2}, \pm 9, \pm \frac{9}{2}, \pm 21, \pm \frac{21}{2}, \pm 63, \pm \frac{63}{2}$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$2 {(3)}^{3} - 7 {(3)}^{2} - 18 \cdot 3 + 63$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 3$ is a root of the polynomial.

Simplify each term.

Raise $3$ to the power of $3$ .

$2 \cdot 27 - 7 {(3)}^{2} - 18 \cdot 3 + 63$

Multiply $2$ by $27$ .

$54 - 7 {(3)}^{2} - 18 \cdot 3 + 63$

Raise $3$ to the power of $2$ .

$54 - 7 \cdot 9 - 18 \cdot 3 + 63$

Multiply $- 7$ by $9$ .

$54 - 63 - 18 \cdot 3 + 63$

Multiply $- 18$ by $3$ .

$54 - 63 - 54 + 63$

Simplify by adding and subtracting.

Subtract $63$ from $54$ .

$- 9 - 54 + 63$

Subtract $54$ from $- 9$ .

$- 63 + 63$

Add $- 63$ and $63$ .

$0$

Step 5

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} - 7 x^{2} - 18 x + 63}{x - 3}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Place the numbers representing the divisor and the dividend into a division-like configuration.

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0			\xa0

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(3)$ and place the result of $(6)$ under the next term in the dividend $(- 7)$ .

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0	$6$		\xa0
	$2$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0	$6$		\xa0
	$2$	$- 1$

Multiply the newest entry in the result $(- 1)$ by the divisor $(3)$ and place the result of $(- 3)$ under the next term in the dividend $(- 18)$ .

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0	$6$	$- 3$	\xa0
	$2$	$- 1$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0	$6$	$- 3$	\xa0
	$2$	$- 1$	$- 21$

Multiply the newest entry in the result $(- 21)$ by the divisor $(3)$ and place the result of $(- 63)$ under the next term in the dividend $(63)$ .

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0	$6$	$- 3$	$- 63$
	$2$	$- 1$	$- 21$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$3$	$2$	$- 7$	$- 18$	$63$
	\xa0	$6$	$- 3$	$- 63$
	$2$	$- 1$	$- 21$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$2 x^{2} + - 1 x - 21$

Simplify the quotient polynomial.

$2 x^{2} - x - 21$

Step 7

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 21 = - 42$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$(x - 3) + 2 x^{2} - (x) - 21$

Rewrite $- 1$ as $6$ plus $- 7$

$(x - 3) + 2 x^{2} + (6 - 7) x - 21$

Apply the distributive property.

$(x - 3) + 2 x^{2} + 6 x - 7 x - 21$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 3) (2 x^{2} + 6 x) - 7 x - 21$

Factor out the greatest common factor (GCF) from each group.

$(x - 3) + 2 x (x + 3) - 7 (x + 3)$

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x - 3) + (x + 3) (2 x - 7)$

$(x - 3) (x + 3) (2 x - 7)$

Step 8

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{3} - 7 x^{2}) - 18 x + 63 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (2 x - 7) - 9 (2 x - 7) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 7$ .

$(2 x - 7) (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(2 x - 7) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$(2 x - 7) (x + 3) (x - 3) = 0$

Step 9

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 7 = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 10

Set $2 x - 7$ equal to $0$ and solve for $x$ .

Set $2 x - 7$ equal to $0$ .

$2 x - 7 = 0$

Solve $2 x - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

Step 11

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 12

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 13

The final solution is all the values that make $(2 x - 7) (x + 3) (x - 3) = 0$ true.

$x = \frac{7}{2}, - 3, 3$

Step 14

Find the Roots/Zeros Using the Rational Roots Test 2x^3-7x^2-18x+63

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