Solve (x+a)(x+b)=x^2+ax+bx+ab

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

Solve for a (complex solution)

$$a\in \mathrm{C}$$

Solution Steps

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

$$x^{2}+xb+ax+ab=x^{2}+ax+bx+ab$$

Subtract ax from both sides.

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

Combine ax and -ax to get 0.

$$x^{2}+xb+ab=x^{2}+bx+ab$$

Subtract ab from both sides.

$$x^{2}+xb+ab-ab=x^{2}+bx$$

Combine ab and -ab to get 0.

$$x^{2}+xb=x^{2}+bx$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{C}$$

Solve for b (complex solution)

$$b\in \mathrm{C}$$

Solution Steps

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

$$x^{2}+xb+ax+ab=x^{2}+ax+bx+ab$$

Subtract bx from both sides.

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

Combine xb and -bx to get 0.

$$x^{2}+ax+ab=x^{2}+ax+ab$$

Subtract ab from both sides.

$$x^{2}+ax+ab-ab=x^{2}+ax$$

Combine ab and -ab to get 0.

$$x^{2}+ax=x^{2}+ax$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{C}$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

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

$$x^{2}+xb+ax+ab=x^{2}+ax+bx+ab$$

Subtract ax from both sides.

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

Combine ax and -ax to get 0.

$$x^{2}+xb+ab=x^{2}+bx+ab$$

Subtract ab from both sides.

$$x^{2}+xb+ab-ab=x^{2}+bx$$

Combine ab and -ab to get 0.

$$x^{2}+xb=x^{2}+bx$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\in \mathrm{R}$$

Solution Steps

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

$$x^{2}+xb+ax+ab=x^{2}+ax+bx+ab$$

Subtract bx from both sides.

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

Combine xb and -bx to get 0.

$$x^{2}+ax+ab=x^{2}+ax+ab$$

Subtract ab from both sides.

$$x^{2}+ax+ab-ab=x^{2}+ax$$

Combine ab and -ab to get 0.

$$x^{2}+ax=x^{2}+ax$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$