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question 16
solve the equation 5t^3 + 5t^2 - 45t - 45 = 0 for t using algebraic techniques. select all that apply.
t = - 3
nothing in this list is correct.
t = 1
t = - 1
t = - 9
t = 3

Explanation:

Step1: Factor by grouping

Group the terms of the equation $5t^{3}+5t^{2}-45t - 45 = 0$.
We get $5t^{2}(t + 1)-45(t + 1)=0$.
Then factor out $(t + 1)$: $(t + 1)(5t^{2}-45)=0$.

Step2: Factor out common factor from second - factor

Factor out 5 from $5t^{2}-45$: $(t + 1)\times5(t^{2}-9)=0$.
Since $t^{2}-9=(t + 3)(t - 3)$ (using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ where $a=t$ and $b = 3$), the equation becomes $5(t + 1)(t + 3)(t - 3)=0$.

Step3: Solve for $t$

Set each factor equal to zero:
If $t+1 = 0$, then $t=-1$.
If $t + 3=0$, then $t=-3$.
If $t - 3=0$, then $t = 3$.

Answer:

$t=-3$, $t=-1$, $t = 3$