Solve (3a+x)(3a-x)+(2a+x)^2

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

Evaluate

$$a\left(4x+13a\right)$$

Solution Steps

Consider \left(3a+x\right)\left(3a-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(3a\right)^{2}-x^{2}+\left(2a+x\right)^{2}$$

Expand \left(3a\right)^{2}.

$$3^{2}a^{2}-x^{2}+\left(2a+x\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9a^{2}-x^{2}+\left(2a+x\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+x\right)^{2}.

$$9a^{2}-x^{2}+4a^{2}+4ax+x^{2}$$

Combine 9a^{2} and 4a^{2} to get 13a^{2}.

$$13a^{2}-x^{2}+4ax+x^{2}$$

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

$$13a^{2}+4ax$$

Expand

$$4ax+13a^{2}$$

Solution Steps

Consider \left(3a+x\right)\left(3a-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(3a\right)^{2}-x^{2}+\left(2a+x\right)^{2}$$

Expand \left(3a\right)^{2}.

$$3^{2}a^{2}-x^{2}+\left(2a+x\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9a^{2}-x^{2}+\left(2a+x\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+x\right)^{2}.

$$9a^{2}-x^{2}+4a^{2}+4ax+x^{2}$$

Combine 9a^{2} and 4a^{2} to get 13a^{2}.

$$13a^{2}-x^{2}+4ax+x^{2}$$

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

$$13a^{2}+4ax$$