Multiply $ \color{blue}{5} $ by $ \left( a-7\right) $ $$ \color{blue}{5} \cdot \left( a-7\right) = 5a-35 $$

Find $ \left(b+2\right)^2 $ using formula.

$$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ b } $ and $ B = \color{red}{ 2 }$.

$$ \begin{aligned}\left(b+2\right)^2 = \color{blue}{b^2} +2 \cdot b \cdot 2 + \color{red}{2^2} = b^2+4b+4\end{aligned} $$

Multiply $5a-35$ by $ \dfrac{b^2+4b+4}{20} $ to get $ \dfrac{5ab^2+20ab-35b^2+20a-140b-140}{20} $.

Step 1: Write $ 5a-35 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 5a-35 \cdot \frac{b^2+4b+4}{20} & \xlongequal{\text{Step 1}} \frac{5a-35}{\color{red}{1}} \cdot \frac{b^2+4b+4}{20} \xlongequal{\text{Step 2}} \frac{ \left( 5a-35 \right) \cdot \left( b^2+4b+4 \right) }{ 1 \cdot 20 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5ab^2+20ab+20a-35b^2-140b-140 }{ 20 } = \frac{5ab^2+20ab-35b^2+20a-140b-140}{20} \end{aligned} $$

Multiply $ \dfrac{5ab^2+20ab-35b^2+20a-140b-140}{20} $ by $ a $ to get $ \dfrac{ 5a^2b^2+20a^2b-35ab^2+20a^2-140ab-140a }{ 20 } $.

Step 1: Write $ a $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5ab^2+20ab-35b^2+20a-140b-140}{20} \cdot a & \xlongequal{\text{Step 1}} \frac{5ab^2+20ab-35b^2+20a-140b-140}{20} \cdot \frac{a}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 5ab^2+20ab-35b^2+20a-140b-140 \right) \cdot a }{ 20 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 5a^2b^2+20a^2b-35ab^2+20a^2-140ab-140a }{ 20 } \end{aligned} $$

Multiply $ \dfrac{5a^2b^2+20a^2b-35ab^2+20a^2-140ab-140a}{20} $ by $ 7-a $ to get $ \dfrac{-5a^3b^2-20a^3b+70a^2b^2-20a^3+280a^2b-245ab^2+280a^2-980ab-980a}{20} $.

Step 1: Write $ 7-a $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5a^2b^2+20a^2b-35ab^2+20a^2-140ab-140a}{20} \cdot 7-a & \xlongequal{\text{Step 1}} \frac{5a^2b^2+20a^2b-35ab^2+20a^2-140ab-140a}{20} \cdot \frac{7-a}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 5a^2b^2+20a^2b-35ab^2+20a^2-140ab-140a \right) \cdot \left( 7-a \right) }{ 20 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 35a^2b^2-5a^3b^2+140a^2b-20a^3b-245ab^2+35a^2b^2+140a^2-20a^3-980ab+140a^2b-980a+140a^2 }{ 20 } = \\[1ex] &= \frac{-5a^3b^2-20a^3b+70a^2b^2-20a^3+280a^2b-245ab^2+280a^2-980ab-980a}{20} \end{aligned} $$

Multiply $ \dfrac{-5a^3b^2-20a^3b+70a^2b^2-20a^3+280a^2b-245ab^2+280a^2-980ab-980a}{20} $ by $ b+2 $ to get $ \dfrac{-5a^3b^3-30a^3b^2+70a^2b^3-60a^3b+420a^2b^2-245ab^3-40a^3+840a^2b-1470ab^2+560a^2-2940ab-1960a}{20} $.

Step 1: Write $ b+2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{-5a^3b^2-20a^3b+70a^2b^2-20a^3+280a^2b-245ab^2+280a^2-980ab-980a}{20} \cdot b+2 & \xlongequal{\text{Step 1}} \frac{-5a^3b^2-20a^3b+70a^2b^2-20a^3+280a^2b-245ab^2+280a^2-980ab-980a}{20} \cdot \frac{b+2}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( -5a^3b^2-20a^3b+70a^2b^2-20a^3+280a^2b-245ab^2+280a^2-980ab-980a \right) \cdot \left( b+2 \right) }{ 20 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -5a^3b^3-10a^3b^2-20a^3b^2-40a^3b+70a^2b^3+140a^2b^2-20a^3b-40a^3+280a^2b^2+560a^2b-245ab^3-490ab^2+280a^2b+560a^2-980ab^2-1960ab-980ab-1960a }{ 20 } = \\[1ex] &= \frac{-5a^3b^3-30a^3b^2+70a^2b^3-60a^3b+420a^2b^2-245ab^3-40a^3+840a^2b-1470ab^2+560a^2-2940ab-1960a}{20} \end{aligned} $$