Solve (5a+3b)^2+(-a+2b)^2

$$( 5 a + 3 b ) ^ { 2 } + ( - a + 2 b ) ^ { 2 }$$

Evaluate

$$13\left(2a^{2}+2ab+b^{2}\right)$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(5a+3b\right)^{2}.

$$25a^{2}+30ab+9b^{2}+\left(-a+2b\right)^{2}$$

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

$$25a^{2}+30ab+9b^{2}+\left(-a\right)^{2}+4\left(-a\right)b+4b^{2}$$

Calculate -a to the power of 2 and get a^{2}.

$$25a^{2}+30ab+9b^{2}+a^{2}+4\left(-a\right)b+4b^{2}$$

Combine 25a^{2} and a^{2} to get 26a^{2}.

$$26a^{2}+30ab+9b^{2}+4\left(-a\right)b+4b^{2}$$

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

$$26a^{2}+30ab+13b^{2}+4\left(-a\right)b$$

Multiply 4 and -1 to get -4.

$$26a^{2}+30ab+13b^{2}-4ab$$

Combine 30ab and -4ab to get 26ab.

$$26a^{2}+26ab+13b^{2}$$

Expand

$$26a^{2}+26ab+13b^{2}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(5a+3b\right)^{2}.

$$25a^{2}+30ab+9b^{2}+\left(-a+2b\right)^{2}$$

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

$$25a^{2}+30ab+9b^{2}+\left(-a\right)^{2}+4\left(-a\right)b+4b^{2}$$

Calculate -a to the power of 2 and get a^{2}.

$$25a^{2}+30ab+9b^{2}+a^{2}+4\left(-a\right)b+4b^{2}$$

Combine 25a^{2} and a^{2} to get 26a^{2}.

$$26a^{2}+30ab+9b^{2}+4\left(-a\right)b+4b^{2}$$

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

$$26a^{2}+30ab+13b^{2}+4\left(-a\right)b$$

Multiply 4 and -1 to get -4.

$$26a^{2}+30ab+13b^{2}-4ab$$

Combine 30ab and -4ab to get 26ab.

$$26a^{2}+26ab+13b^{2}$$