Find the Roots/Zeros Using the Rational Roots Test 7x^3-5x^2-63x+45

$7 x^{3} - 5 x^{2} - 63 x + 45$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

$q = \pm 1, \pm 7$

Step 2

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

$\pm 1, \pm \frac{1}{7}, \pm 3, \pm \frac{3}{7}, \pm 5, \pm \frac{5}{7}, \pm 9, \pm \frac{9}{7}, \pm 15, \pm \frac{15}{7}, \pm 45, \pm \frac{45}{7}$

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.

$7 {(3)}^{3} - 5 {(3)}^{2} - 63 \cdot 3 + 45$

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 $3$ .

$7 \cdot 27 - 5 {(3)}^{2} - 63 \cdot 3 + 45$

Multiply $7$ by $27$ .

$189 - 5 {(3)}^{2} - 63 \cdot 3 + 45$

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

$189 - 5 \cdot 9 - 63 \cdot 3 + 45$

Multiply $- 5$ by $9$ .

$189 - 45 - 63 \cdot 3 + 45$

Multiply $- 63$ by $3$ .

$189 - 45 - 189 + 45$

Simplify by adding and subtracting.

Subtract $45$ from $189$ .

$144 - 189 + 45$

Subtract $189$ from $144$ .

$- 45 + 45$

Add $- 45$ and $45$ .

$0$

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.

$\frac{7 x^{3} - 5 x^{2} - 63 x + 45}{x - 3}$

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.

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0			\xa0

The first number in the dividend $(7)$ is put into the first position of the result area (below the horizontal line).

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0			\xa0
	$7$

Multiply the newest entry in the result $(7)$ by the divisor $(3)$ and place the result of $(21)$ under the next term in the dividend $(- 5)$ .

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0	$21$		\xa0
	$7$

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

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0	$21$		\xa0
	$7$	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(3)$ and place the result of $(48)$ under the next term in the dividend $(- 63)$ .

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0	$21$	$48$	\xa0
	$7$	$16$

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

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0	$21$	$48$	\xa0
	$7$	$16$	$- 15$

Multiply the newest entry in the result $(- 15)$ by the divisor $(3)$ and place the result of $(- 45)$ under the next term in the dividend $(45)$ .

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0	$21$	$48$	$- 45$
	$7$	$16$	$- 15$

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

$3$	$7$	$- 5$	$- 63$	$45$
	\xa0	$21$	$48$	$- 45$
	$7$	$16$	$- 15$	$0$

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

$7 x^{2} + 16 x - 15$

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 $a \cdot c = 7 \cdot - 15 = - 105$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$(x - 3) + 7 x^{2} + 16 (x) - 15$

Rewrite $16$ as $- 5$ plus $21$

$(x - 3) + 7 x^{2} + (- 5 + 21) x - 15$

Apply the distributive property.

$(x - 3) + 7 x^{2} - 5 x + 21 x - 15$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 3) (7 x^{2} - 5 x) + 21 x - 15$

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

$(x - 3) + x (7 x - 5) + 3 (7 x - 5)$

Factor the polynomial by factoring out the greatest common factor, $7 x - 5$ .

$(x - 3) + (7 x - 5) (x + 3)$

$(x - 3) (7 x - 5) (x + 3)$

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.

$(7 x^{3} - 5 x^{2}) - 63 x + 45 = 0$

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

$x^{2} (7 x - 5) - 9 (7 x - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x - 5$ .

$(7 x - 5) (x^{2} - 9) = 0$

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

$(7 x - 5) (x^{2} - 3^{2}) = 0$

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 $b = 3$ .

$(7 x - 5) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$(7 x - 5) (x + 3) (x - 3) = 0$

Step 9

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

$7 x - 5 = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 10

Set $7 x - 5$ equal to $0$ and solve for $x$ .

Set $7 x - 5$ equal to $0$ .

$7 x - 5 = 0$

Solve $7 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$7 x = 5$

Divide each term in $7 x = 5$ by $7$ and simplify.

Divide each term in $7 x = 5$ by $7$ .

$\frac{7 x}{7} = \frac{5}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{5}{7}$

Divide $x$ by $1$ .

$x = \frac{5}{7}$

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 $(7 x - 5) (x + 3) (x - 3) = 0$ true.

$x = \frac{5}{7}, - 3, 3$

Step 14

Find the Roots/Zeros Using the Rational Roots Test 7x^3-5x^2-63x+45

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