Factor the Quadratic

Question

-16t^2+32t+20=0. How are you supposed to find -5 and positive 1 to put in the parenthesis? -4(2t-5)(2t+1)?

Answer 1

$-16t^2+32t+20=-16(t^2-2t-5/4) $. Roots of $t^2-2t-5/4$ are $$\frac{2\pm \sqrt{4+5}}{2} = 1\pm 3/2,$$ hence $$-16t^2+32t+20=-16(t-(1-3/2))(t-(1+3/2))=-16(t-5/2)(t+1/2)=-4(2t-5)(2t+1)$$

Answer 2

You know that $-4(2t-5)(2t+1)=0$.

What values of $t$ will make the above result in $0$? If we think a little, we can note that if any of the terms are being multiplied by zero, then the product is zero as well, since $c\times 0=0$ for any value of $c$.

To determine what values of $t$ would make one of the terms zero we proceed to see how each term could equal zero like so:

We note that if $(2t-5)=0$, then the entire left side of the equation would equal zero since we are multiplying terms by $0$. To solve this, we add $5$ to both sides and divide both sides by $2$ giving us $2t-5=0\implies 2t=5\implies t=\frac{5}{2}$, so this is one value of $t$ that would result in the left hand side of the equation being zero. So this is one of our solutions.

Similarly, we note that if $(2t+1)=0$, then the left hand side of the equation would equal zero. We solve for $t$ in this. We subtract $1$ from both sides and divide by $2$ giving us $2t+1=0\implies 2t=-1\implies t=-\frac{1}{2}$.

Hence the solutions of the quadratic, i.e., the values of $t$ which would make $-4(2t-5)(2t+1)=0$ are $t=\frac{5}{2}$ and $t=-\frac{1}{2}$