We are given the equation:

$4x^3 - 3x^2 + ax + 15 = (4x + 5)(x^2 + bx + 3)$

To find $a + b$, we'll start by expanding the right-hand side.

Step 1: Expand the right-hand side of the equation

Using the distributive property, expand $(4x + 5)(x^2 + bx + 3)$:

$(4x + 5)(x^2 + bx + 3) = 4x(x^2 + bx + 3) + 5(x^2 + bx + 3)$

Now expand each term:

$4x(x^2 + bx + 3) = 4x^3 + 4bx^2 + 12x$ $5(x^2 + bx + 3) = 5x^2 + 5bx + 15$

Now combine all terms:

$4x^3 + 4bx^2 + 12x + 5x^2 + 5bx + 15$

Simplify the expression:

$4x^3 + (4b + 5)x^2 + (12 + 5b)x + 15$

Step 2: Set the expanded equation equal to the given equation

We now have:

$4x^3 - 3x^2 + ax + 15 = 4x^3 + (4b + 5)x^2 + (12 + 5b)x + 15$

Step 3: Compare the coefficients of like terms

• For the $x^3$ terms: $4 = 4 \quad \text{(which is already true)}$

• For the $x^2$ terms: $-3 = 4b + 5$ Solve for $b$: $4b + 5 = -3 \quad \Rightarrow \quad 4b = -8 \quad \Rightarrow \quad b = -2$

• For the $x$ terms: $a = 12 + 5b$ Substitute $b = -2$: $a = 12 + 5(-2) = 12 - 10 = 2$

Step 4: Find $a + b$

Now that we know $a = 2$ and $b = -2$, we can find:

$a + b = 2 + (-2) = 0$

Final Answer:

$a + b = 0$

Would you like a more detailed breakdown, or do you have any specific questions?

Related Questions:

1. How do you expand binomials like $(4x + 5)(x^2 + bx + 3)$?
2. What are the different ways to compare coefficients in polynomial equations?
3. How would you solve for $a$ and $b$ in similar polynomial equations?
4. Can this technique be applied to higher degree polynomials?
5. How do you handle constant terms when comparing polynomials?

Tip:

When comparing coefficients, always pay attention to the powers of $x$ and make sure that each term on both sides corresponds to the same power of $x$.