Solve .1(x-3)^2(x-1)^2

$$.1 ( x - 3 ) ^ { 2 } ( x - 1 ) ^ { 2 }$$

Evaluate

$$\frac{\left(\left(x-3\right)\left(x-1\right)\right)^{2}}{10}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0.1\left(x^{2}-6x+9\right)\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$0.1\left(x^{2}-6x+9\right)\left(x^{2}-2x+1\right)$$

Use the distributive property to multiply 0.1 by x^{2}-6x+9.

$$\left(0.1x^{2}-0.6x+0.9\right)\left(x^{2}-2x+1\right)$$

Use the distributive property to multiply 0.1x^{2}-0.6x+0.9 by x^{2}-2x+1 and combine like terms.

$$0.1x^{4}-0.8x^{3}+2.2x^{2}-2.4x+0.9$$

Expand

$$\frac{x^{4}}{10}-\frac{4x^{3}}{5}+\frac{11x^{2}}{5}-\frac{12x}{5}+0.9$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0.1\left(x^{2}-6x+9\right)\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$0.1\left(x^{2}-6x+9\right)\left(x^{2}-2x+1\right)$$

Use the distributive property to multiply 0.1 by x^{2}-6x+9.

$$\left(0.1x^{2}-0.6x+0.9\right)\left(x^{2}-2x+1\right)$$

Use the distributive property to multiply 0.1x^{2}-0.6x+0.9 by x^{2}-2x+1 and combine like terms.

$$0.1x^{4}-0.8x^{3}+2.2x^{2}-2.4x+0.9$$