Question 86·Medium·Equivalent Expressions

After you expand, write the result in the form $\text{(something)}x^{3} + \text{(something)}x^{2} + \text{(something)}x + \text{(constant)}$ and compare each coefficient with $15x^{3}+14x^{2}-7x-6$.

The $x^{3}$ term and the constant term each involve only one of the unknowns ($a$ or $c$). Find $a$ and $c$ from these first, then plug them into the equation involving $b$.

Once you know $a$, use the equation that comes from the $x^{2}$ coefficients to solve for $b$.

In Desmos, enter the original polynomial as f(x) = 15x^3 + 14x^2 - 7x - 6. Then enter the factored form with sliders as g(x) = (3x - 2)(a x^2 + b x + c); Desmos will create sliders for a, b, and c.

Adjust the sliders for a, b, and c until the graph of g(x) lies exactly on top of the graph of f(x) for all visible $x$-values. When the curves coincide perfectly, the slider values for a, b, and c are the correct coefficients, and you can read off the value of b directly from its slider.

We are told

Here $a$, $b$, and $c$ are constants we need to find. We only care about $b$, but we will first find $a$ and $c$ to make solving for $b$ easier.

Distribute $(3x-2)$ over $(ax^{2}+bx+c)$:

• Multiply by $3x$: $3x(ax^{2}+bx+c) = 3ax^{3} + 3bx^{2} + 3cx$.
• Multiply by $-2$: $-2(ax^{2}+bx+c) = -2ax^{2} - 2bx - 2c$.

Add these together:

Now match coefficients of like powers of $x$ on both sides of

This gives the system:

• From $x^{3}$: $3a = 15$.
• From $x^{2}$: $3b - 2a = 14$.
• From $x^{1}$: $3c - 2b = -7$.
• From the constant term: $-2c = -6$.

We can now solve these simple equations step by step.

Solve the equations that involve only one variable:

• From $3a = 15$, divide both sides by $3$ to get $a = 5$.
• From $-2c = -6$, divide both sides by $-2$ to get $c = 3$.

Now substitute $a = 5$ and later, if needed, $c = 3$ into the other equations to solve for $b$.

Use the $x^{2}$-coefficient equation $3b - 2a = 14$ and substitute $a = 5$:

So, the value of $b$ is $8$.