Find the Roots (Zeros) p(x)=(2x^2-9x+7)(x-2)

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

Set $(2 x^{2} - 9 x + 7) (x - 2)$ equal to $0$ .

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

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 $0$ .

$2 x^{2} - 9 x + 7 = 0$

$x - 2 = 0$

Set $2 x^{2} - 9 x + 7$ equal to $0$ and solve for $x$ .

Set $2 x^{2} - 9 x + 7$ equal to $0$ .

$2 x^{2} - 9 x + 7 = 0$

Solve $2 x^{2} - 9 x + 7 = 0$ 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 $a \cdot c = 2 \cdot 7 = 14$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$2 x^{2} - 9 x + 7 = 0$

Rewrite $- 9$ as $- 2$ plus $- 7$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$2 x (x - 1) - 7 (x - 1) = 0$

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

$(x - 1) (2 x - 7) = 0$

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

$x - 1 = 0$

$2 x - 7 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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 $1$ .

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

The final solution is all the values that make $(x - 1) (2 x - 7) = 0$ true.

$x = 1, \frac{7}{2}$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(2 x^{2} - 9 x + 7) (x - 2) = 0$ true.

$x = 1, \frac{7}{2}, 2$

Find the Roots (Zeros) p(x)=(2x^2-9x+7)(x-2)

Related Questions

Support LCMGCF.com by disabling your adblocker.