Solve g(t)=-1/4(t-3)^2(t+3)^2

$$g ( t ) = - \frac { 1 } { 4 } ( t - 3 ) ^ { 2 } ( t + 3 ) ^ { 2 }$$

Evaluate

$$-\frac{\left(t^{2}-9\right)^{2}}{4}$$

Solution Steps

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

$$-\frac{1}{4}\left(t^{2}-6t+9\right)\left(t+3\right)^{2}$$

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

$$-\frac{1}{4}\left(t^{2}-6t+9\right)\left(t^{2}+6t+9\right)$$

Use the distributive property to multiply -\frac{1}{4} by t^{2}-6t+9.

$$\left(-\frac{1}{4}t^{2}+\frac{3}{2}t-\frac{9}{4}\right)\left(t^{2}+6t+9\right)$$

Use the distributive property to multiply -\frac{1}{4}t^{2}+\frac{3}{2}t-\frac{9}{4} by t^{2}+6t+9 and combine like terms.

$$-\frac{1}{4}t^{4}+\frac{9}{2}t^{2}-\frac{81}{4}$$

Expand

$$-\frac{t^{4}}{4}+\frac{9t^{2}}{2}-\frac{81}{4}$$

Solution Steps

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

$$-\frac{1}{4}\left(t^{2}-6t+9\right)\left(t+3\right)^{2}$$

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

$$-\frac{1}{4}\left(t^{2}-6t+9\right)\left(t^{2}+6t+9\right)$$

Use the distributive property to multiply -\frac{1}{4} by t^{2}-6t+9.

$$\left(-\frac{1}{4}t^{2}+\frac{3}{2}t-\frac{9}{4}\right)\left(t^{2}+6t+9\right)$$

Use the distributive property to multiply -\frac{1}{4}t^{2}+\frac{3}{2}t-\frac{9}{4} by t^{2}+6t+9 and combine like terms.

$$-\frac{1}{4}t^{4}+\frac{9}{2}t^{2}-\frac{81}{4}$$