Find the Holes in the Graph f(x)=(2x^2+5x-3)/(x+3)

$f (x) = \frac{2 x^{2} + 5 x - 3}{x + 3}$

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$f (x) = \frac{2 x^{2} + 5 (x) - 3}{x + 3}$

Rewrite $5$ as $- 1$ plus $6$

$f (x) = \frac{2 x^{2} + (- 1 + 6) x - 3}{x + 3}$

Apply the distributive property.

$f (x) = \frac{2 x^{2} - 1 x + 6 x - 3}{x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(2 x^{2} - 1 x) + 6 x - 3}{x + 3}$

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

$f (x) = \frac{x (2 x - 1) + 3 (2 x - 1)}{x + 3}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$f (x) = \frac{(2 x - 1) (x + 3)}{x + 3}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$f (x) = \frac{(2 x - 1) (x + 3)}{x + 3}$

Divide $2 x - 1$ by $1$ .

$f (x) = 2 x - 1$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x + 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $2 x - 1$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $2 x - 1$ and simplify.

Substitute $- 3$ for $x$ to find the $y$ coordinate of the hole.

$2 \cdot - 3 - 1$

Simplify.

Multiply $2$ by $- 3$ .

$- 6 - 1$

Subtract $1$ from $- 6$ .

$- 7$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(- 3, - 7)$

Find the Holes in the Graph f(x)=(2x^2+5x-3)/(x+3)

Related Questions

Support LCMGCF.com by disabling your adblocker.