Solve for d x^2+2x-15=(ax+b)(cx+d)

$x^{2} + 2 x - 15 = (a x + b) (c x + d)$

Step 1

Rewrite the equation as $(a x + b) (c x + d) = x^{2} + 2 x - 15$ .

$(a x + b) (c x + d) = x^{2} + 2 x - 15$

Step 2

Divide each term in $(a x + b) (c x + d) = x^{2} + 2 x - 15$ by $a x + b$ and simplify.

Divide each term in $(a x + b) (c x + d) = x^{2} + 2 x - 15$ by $a x + b$ .

$\frac{(a x + b) (c x + d)}{a x + b} = \frac{x^{2}}{a x + b} + \frac{2 x}{a x + b} + \frac{- 15}{a x + b}$

Simplify the left side.

Cancel the common factor of $a x + b$ .

Cancel the common factor.

$\frac{(a x + b) (c x + d)}{a x + b} = \frac{x^{2}}{a x + b} + \frac{2 x}{a x + b} + \frac{- 15}{a x + b}$

Divide $c x + d$ by $1$ .

$c x + d = \frac{x^{2}}{a x + b} + \frac{2 x}{a x + b} + \frac{- 15}{a x + b}$

Simplify the right side.

Combine into one fraction.

Move the negative in front of the fraction.

$c x + d = \frac{x^{2}}{a x + b} + \frac{2 x}{a x + b} - \frac{15}{a x + b}$

Combine the numerators over the common denominator.

$c x + d = \frac{x^{2} + 2 x}{a x + b} - \frac{15}{a x + b}$

Combine the numerators over the common denominator.

$c x + d = \frac{x^{2} + 2 x - 15}{a x + b}$

Factor $x^{2} + 2 x - 15$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$c x + d = \frac{(x - 3) (x + 5)}{a x + b}$

Step 3

Subtract $c x$ from both sides of the equation.

$d = \frac{(x - 3) (x + 5)}{a x + b} - c x$

Solve for d x^2+2x-15=(ax+b)(cx+d)

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