Case I.

106. When all the terms have a common factor.

1. Resolve into factors $3a^2-6ab$.

Since $3a$ is seen to be a factor of each term, we have

$$\begin{gathered} \frac{3a^2-6ab}{3a}\&=\frac{3a^2}{3a}-\frac{6ab}{3a}=a-2b. \\ \therefore 3a^2-6ab\&=3a(a-2b). \end{gathered}$$

Hence, the required factors are $3a$ and $a-2b$.

2. Resolve into factors $4x^3+12x^2-8x$.

Since $4x$ is seen to be a factor of each term, we have

$$\begin{gathered} \frac{4x^3+12x^2-8x}{4x}\&=\frac{4x^3}{4x}+\frac{12x^2}{4x}-\frac{8x}{4x} \\ \&=x^2+3x-2. \\ \therefore 4x^3+12x^2-8x\&=4x(x^2+3x-2). \end{gathered}$$

Hence the required factors are $4x$ and $x^2+3x-2$.

Exercise 31.

Resolve into two factors:

1. $2x^2-4x$.

2. $3a^3-6a$.

3. $5a^2{b}^2-10a^3{b}^3$.

4. $3x^{2}y+4x{y}^2$.

5. $8a^3{b}^2+4a^2{b}^3$.

6. $3a^4-12a^2-6a^3$.

7. $4x^2-8x^4-12x^5$.

8. $5-10x^2{y}^2+15x^{2}y$.

9. $7a^2+14a-21a^3$.

10. $3x^3{y}^3-6x^4{y}^4-9x^2{y}^2$.