Question 197·Medium·Equivalent Expressions

Think of $-(x^2 - 7x - 2)$ as multiplying the whole parentheses by $-1$. How does that change the signs of each term inside?

Once both parentheses are expanded, group $x^2$ terms together, $x$ terms together, and constants together, and then add them.

In Desmos, type f(x) = 2(3x^2 - 5x + 4) - (x^2 - 7x - 2) so you can compare every choice to this expression.

Type each option as its own function, for example:

• g(x) = 7x^2 - 17x + 6
• h(x) = 5x^2 - 3x + 10
• p(x) = 5x^2 - 17x + 6
• q(x) = 7x^2 - 3x + 10 Then use a table for each (click the gear icon, then the "+" and choose "table") to compare values of $f(x)$ and each option at several $x$-values (like $x=0,1,2$).

Look for the function whose table values match $f(x)$ for every $x$ you test; that function’s expression is the one equivalent to the original.

Start by expanding $2(3x^2 - 5x + 4)$.

Multiply each term inside the parentheses by 2:

• $2 \cdot 3x^2 = 6x^2$
• $2 \cdot (-5x) = -10x$
• $2 \cdot 4 = 8$

So the first part becomes $6x^2 - 10x + 8$.

Now deal with $-(x^2 - 7x - 2)$.

A minus sign in front of parentheses means you multiply every term inside by $-1$:

• $-(x^2) = -x^2$
• $-(-7x) = +7x$
• $-(-2) = +2$

So $-(x^2 - 7x - 2)$ becomes $-x^2 + 7x + 2$.

Substitute the expanded pieces back into the original expression: $2(3x^2 - 5x + 4) - (x^2 - 7x - 2)$ becomes $(6x^2 - 10x + 8) + (-x^2 + 7x + 2)$. Now you are adding two polynomials: $6x^2 - 10x + 8$ and $-x^2 + 7x + 2$.

Combine like terms:

• $x^2$ terms: $6x^2 + (-x^2) = 5x^2$
• $x$ terms: $-10x + 7x = -3x$
• Constant terms: $8 + 2 = 10$

So the simplified expression is $5x^2 - 3x + 10$, which matches answer choice B.