Find the Roots (Zeros) (2x^2+7x+5)(x-3)

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

Step 1

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

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

Step 2

Solve for $x$ .

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

$2 x^{2} + 7 x + 5 = 0$

$x - 3 = 0$

Set $2 x^{2} + 7 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x^{2} + 7 x + 5$ equal to $0$ .

$2 x^{2} + 7 x + 5 = 0$

Solve $2 x^{2} + 7 x + 5 = 0$ for $x$ .

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 $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = 7$ .

Factor $7$ out of $7 x$ .

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

Rewrite $7$ as $2$ plus $5$

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

Apply the distributive property.

$2 x^{2} + 2 x + 5 x + 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$2 x (x + 1) + 5 (x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x + 1) (2 x + 5) = 0$

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

$x + 1 = 0$

$2 x + 5 = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

The final solution is all the values that make $(x + 1) (2 x + 5) = 0$ true.

$x = - 1, - \frac{5}{2}$

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

$x = - 1, - \frac{5}{2}, 3$

Step 3

Find the Roots (Zeros) (2x^2+7x+5)(x-3)

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