Find the Roots (Zeros) f(x)=7x^3-5x^2-63x+45

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

Step 1

Set $7 x^{3} - 5 x^{2} - 63 x + 45$ equal to $0$ .

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

Step 2

Solve for $x$ .

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.

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

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

$7 x - 5 = 0$

$x + 3 = 0$

$x - 3 = 0$

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 $1$ .

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

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.

$x = - 3$

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.

$x = 3$

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 3

Find the Roots (Zeros) f(x)=7x^3-5x^2-63x+45

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