Solve x^2+3x-10=(x+a)(x-b)

$$x ^ { 2 } + 3 x - 10 = ( x + a ) ( x - b )$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{10-3x-bx}{x-b}\text{, }&x\neq b\\a\in \mathrm{C}\text{, }&\left(x=2\text{ and }b=2\right)\text{ or }\left(x=-5\text{ and }b=-5\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+a by x-b.

$$x^{2}+3x-10=x^{2}-xb+ax-ab$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-xb+ax-ab=x^{2}+3x-10$$

Subtract x^{2} from both sides.

$$-xb+ax-ab=x^{2}+3x-10-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-xb+ax-ab=3x-10$$

Add xb to both sides.

$$ax-ab=3x-10+xb$$

Combine all terms containing a.

$$\left(x-b\right)a=3x-10+xb$$

The equation is in standard form.

$$\left(x-b\right)a=bx+3x-10$$

Divide both sides by x-b.

$$\frac{\left(x-b\right)a}{x-b}=\frac{bx+3x-10}{x-b}$$

Dividing by x-b undoes the multiplication by x-b.

$$a=\frac{bx+3x-10}{x-b}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{ax-3x+10}{x+a}\text{, }&x\neq -a\\b\in \mathrm{C}\text{, }&\left(x=2\text{ and }a=-2\right)\text{ or }\left(x=-5\text{ and }a=5\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+a by x-b.

$$x^{2}+3x-10=x^{2}-xb+ax-ab$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-xb+ax-ab=x^{2}+3x-10$$

Subtract x^{2} from both sides.

$$-xb+ax-ab=x^{2}+3x-10-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-xb+ax-ab=3x-10$$

Subtract ax from both sides.

$$-xb-ab=3x-10-ax$$

Combine all terms containing b.

$$\left(-x-a\right)b=3x-10-ax$$

The equation is in standard form.

$$\left(-x-a\right)b=-ax+3x-10$$

Divide both sides by -x-a.

$$\frac{\left(-x-a\right)b}{-x-a}=\frac{-ax+3x-10}{-x-a}$$

Dividing by -x-a undoes the multiplication by -x-a.

$$b=\frac{-ax+3x-10}{-x-a}$$

Divide -xa-10+3x by -x-a.

$$b=-\frac{-ax+3x-10}{x+a}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{10-3x-bx}{x-b}\text{, }&x\neq b\\a\in \mathrm{R}\text{, }&\left(x=2\text{ and }b=2\right)\text{ or }\left(x=-5\text{ and }b=-5\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+a by x-b.

$$x^{2}+3x-10=x^{2}-xb+ax-ab$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-xb+ax-ab=x^{2}+3x-10$$

Subtract x^{2} from both sides.

$$-xb+ax-ab=x^{2}+3x-10-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-xb+ax-ab=3x-10$$

Add xb to both sides.

$$ax-ab=3x-10+xb$$

Combine all terms containing a.

$$\left(x-b\right)a=3x-10+xb$$

The equation is in standard form.

$$\left(x-b\right)a=bx+3x-10$$

Divide both sides by x-b.

$$\frac{\left(x-b\right)a}{x-b}=\frac{bx+3x-10}{x-b}$$

Dividing by x-b undoes the multiplication by x-b.

$$a=\frac{bx+3x-10}{x-b}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{ax-3x+10}{x+a}\text{, }&x\neq -a\\b\in \mathrm{R}\text{, }&\left(x=2\text{ and }a=-2\right)\text{ or }\left(x=-5\text{ and }a=5\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+a by x-b.

$$x^{2}+3x-10=x^{2}-xb+ax-ab$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-xb+ax-ab=x^{2}+3x-10$$

Subtract x^{2} from both sides.

$$-xb+ax-ab=x^{2}+3x-10-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-xb+ax-ab=3x-10$$

Subtract ax from both sides.

$$-xb-ab=3x-10-ax$$

Combine all terms containing b.

$$\left(-x-a\right)b=3x-10-ax$$

The equation is in standard form.

$$\left(-x-a\right)b=-ax+3x-10$$

Divide both sides by -x-a.

$$\frac{\left(-x-a\right)b}{-x-a}=\frac{-ax+3x-10}{-x-a}$$

Dividing by -x-a undoes the multiplication by -x-a.

$$b=\frac{-ax+3x-10}{-x-a}$$

Divide 3x-10-ax by -x-a.

$$b=-\frac{-ax+3x-10}{x+a}$$