Question 78·Hard·Equivalent Expressions

After distributing, collect all the $x^2$ terms, all the $x$ terms, and all the constant terms so the expression looks like $ax^2 + cx + d$.

Set your simplified expression equal to $6x^2 + 14x - 8$ and match the coefficient of $x$ and the constant term. What equations in $b$ do you get from these matches?

Solve the equation that comes from the $x$-coefficient (or from the constant term). Make sure the value you find makes both the $x$-coefficient and the constant match.

In Desmos, type f(x) = 6x^2 + b(4 - x) - 3x(b - 2) and accept the prompt to add a slider for b. Then type g(x) = 6x^2 + 14x - 8.

Adjust the slider for b until the graph of f(x) lies exactly on top of the graph of g(x) for all visible $x$-values. The value of b at that point is the solution.

You can also type h(x) = f(x) - g(x) and vary b; the correct value of b will make h(x) equal to $0$ for all $x$ (the graph becomes the $x$-axis). Read that value of b from the slider.

Start with the expression:

Distribute $b$ over $(4 - x)$ and $-3x$ over $(b - 2)$:

• $b(4 - x) = 4b - bx$
• $-3x(b - 2) = -3bx + 6x$

So the whole expression becomes

Group the $x$ terms and constant terms together:

• The $x$ terms are $-bx - 3bx + 6x = (-4b + 6)x$.
• The constant term is $4b$.

So the expression simplifies to

We are told this expression is equivalent to

For two polynomials in $x$ to be equal for all $x$, coefficients of matching powers of $x$ must be equal. The $6x^2$ term already matches, so focus on the $x$ term and the constant term:

• $x$-coefficient: $6 - 4b$ must equal $14$.
• Constant term: $4b$ must equal $-8$.

Either of these equations can be used to solve for $b$; both should lead to the same value.

Use the $x$-coefficient equation:

Subtract $6$ from both sides:

Divide both sides by $-4$:

Check with the constant term: $4b = 4(-2) = -8$, which matches the constant in $6x^2 + 14x - 8$. Therefore, the value of $b$ is $-2$.