Question 166·Easy·Equivalent Expressions

The pattern $ax + bx = (a + b)x$ works in reverse too: if two terms share a factor, you can factor it out. How does this apply to $5t(t - 2) - 3(t - 2)$?

Once you factor the expression, check which option has the same factors (in any order). If expanding helps, you can quickly expand a choice to see if it matches the original expression.

In a Desmos expression line, type f(t) = 5*t*(t - 2) - 3*(t - 2) so you have a function representing the original expression.

On new lines, enter:

• A(t) = (t - 2)*(5*t - 3)
• B(t) = (t - 2)*(5*t + 3)
• C(t) = (t + 2)*(5*t - 3)
• D(t) = 2*(t - 2)*(5*t - 3)

Create a table for t values (for example, $t = -2, -1, 0, 1, 2, 3$) and add columns for f(t), A(t), B(t), C(t), and D(t). The choice whose values always match f(t) for all tested $t$ is the expression equivalent to the original.

Start by distributing in each term to see the simplified polynomial:

So the original expression is equivalent to $5t^2 - 13t + 6$.

Now factor $5t^2 - 13t + 6$.

You need two numbers that:

• multiply to $5 \cdot 6 = 30$, and
• add to $-13$.

Those numbers are $-10$ and $-3$, so split the middle term:

Next, group the terms to prepare for factoring by grouping:

Factor each group:

Now factor out the common binomial $(t - 2)$:

So the original expression is equivalent to $(t - 2)(5t - 3)$, which corresponds to choice A.