Solve (a-2)^2+(1/4a^2+a+3)^2

$$( a - 2 ) ^ { 2 } + ( \frac { 1 } { 4 } a ^ { 2 } + a + 3 ) ^ { 2 }$$

Evaluate

$$\frac{\left(a^{2}+4\right)\left(a^{2}+8a+52\right)}{16}$$

Solution Steps

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

$$a^{2}-4a+4+\left(\frac{1}{4}a^{2}+a+3\right)^{2}$$

Square \frac{1}{4}a^{2}+a+3.

$$a^{2}-4a+4+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}+\frac{5}{2}a^{2}+6a+9$$

Combine a^{2} and \frac{5}{2}a^{2} to get \frac{7}{2}a^{2}.

$$\frac{7}{2}a^{2}-4a+4+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}+6a+9$$

Combine -4a and 6a to get 2a.

$$\frac{7}{2}a^{2}+2a+4+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}+9$$

Add 4 and 9 to get 13.

$$\frac{7}{2}a^{2}+2a+13+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}$$

Expand

$$\frac{a^{4}}{16}+\frac{a^{3}}{2}+\frac{7a^{2}}{2}+2a+13$$

Solution Steps

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

$$a^{2}-4a+4+\left(\frac{1}{4}a^{2}+a+3\right)^{2}$$

Square \frac{1}{4}a^{2}+a+3.

$$a^{2}-4a+4+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}+\frac{5}{2}a^{2}+6a+9$$

Combine a^{2} and \frac{5}{2}a^{2} to get \frac{7}{2}a^{2}.

$$\frac{7}{2}a^{2}-4a+4+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}+6a+9$$

Combine -4a and 6a to get 2a.

$$\frac{7}{2}a^{2}+2a+4+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}+9$$

Add 4 and 9 to get 13.

$$\frac{7}{2}a^{2}+2a+13+\frac{1}{16}a^{4}+\frac{1}{2}a^{3}$$