Factoring equations?

Question

Below, there's a question that I found in my Calculus textbook. I'm probably going to ask the professor tomorrow, but I figured I might want to try and ask you guys, since it doesn't say how to do this in the book.

Anyway, how are you supposed to factor equations like in (a)?

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The height of a projectile fired in the air vertically with initial velocity 25 m/s is

$h(t) = 25t − 4.9t^2$

(a) Compute h(1). [h(1) = 20.1] Show that $h(t) − h(1)$ can be factored with (t − 1) as a factor.

(b) Using part (a), show that the average velocity over the interval $[1, t]$ is $20.1 − 4.9t$ .

(c) Use this formula to find the average velocity over several intervals $[1, t]$ with t close to 1. Then estimate the instantaneous velocity at time $t = 1$.

Answer 1

When you plug in $t=1$ into the equation, the expression simplifies to $-4.9t^2+25t-20.1$. This is factorable (Discriminant is a perfect square) and in fact, when you plug in $t=1$, you obtain $0$, indicating that $(t-1)$ indeed is a factor. You may proceed to long division to find the other factor. But can you try this first?

Answer 2

your term can be written as $$-\frac{1}{10}(t-1)(49t-201)$$ this can be obtained by dividing $$-4.9t^2+25t-20.1$$ by $t-1$

Answer 3

Just do it:

$h(t) - h(1) =$

$25t - 4.9t^2 - (25(1) - 4.9(1)^2)=$

$-4.9t^2 + 25t - 20.1=$

$-4.9t(t-1) - 4.9t + 25t - 20.1 = $

$-4.9t(t-1) + 20.1t - 20.1 = $

$-4.9t(t-1) + 20.1(t-1) =$

$(t-1)(-4.9t + 20.1)$