We are given that $4x^2 - 4x - 15$ is a factor of $4x^3 + ax^2 + bx + 30$. Using the factor theorem, the polynomial $4x^3 + ax^2 + bx + 30$ can be expressed as:

$(4x^2 - 4x - 15)(x + c)$

where $c$ is a constant to be determined. Expanding this product, we match coefficients with the original polynomial to find $a$ and $b$.

Step 1: Expand the product

Expanding $(4x^2 - 4x - 15)(x + c)$, we have:

$(4x^2 - 4x - 15)(x + c) = 4x^3 - 4x^2 - 15x + 4cx^2 - 4cx - 15c$

Combining like terms:

$= 4x^3 + (4c - 4)x^2 + (-4c - 15)x - 15c$

Step 2: Compare coefficients

The original polynomial is $4x^3 + ax^2 + bx + 30$. Equating coefficients, we have:

1. Coefficient of $x^2$: $4c - 4 = a$
2. Coefficient of $x$: $-4c - 15 = b$
3. Constant term: $-15c = 30$

Step 3: Solve for $c$

From the constant term:

$-15c = 30 \implies c = -2$

Step 4: Solve for $a$ and $b$

Using $c = -2$:

1. $a = 4c - 4 = 4(-2) - 4 = -8 - 4 = -12$
2. $b = -4c - 15 = -4(-2) - 15 = 8 - 15 = -7$

Final Answer:

$a = -12, \quad b = -7$

Let me know if you need further clarification! 😊

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Related Questions:

1. What is the Factor Theorem, and how does it apply to polynomial division?
2. How would you confirm that $4x^2 - 4x - 15$ is a factor of the given polynomial?
3. Could the value of $c$ be something other than $-2$? Why or why not?
4. Can you generalize this method to factorize higher-degree polynomials?
5. What happens if the polynomial has irrational or complex coefficients?

Tip:

When solving polynomial factorization problems, always match coefficients systematically—it reduces errors!