Question: Find the quotient and remainder for
$$\frac{-8x^4 + 22x^3 - 25x + 9}{4x - 3}$$
The quotient is:
The remainder is:
$(4x - 3)$ is a factor of $-8x^4 + 22x^3 - 25x + 9$:
True
False
1. **State the problem:** We want to divide the polynomial $$-8x^4 + 22x^3 - 25x + 9$$ by the binomial $$4x - 3$$ and find the quotient and remainder.
2. **Recall the division formula:** For polynomials, $$\text{Dividend} = (\text{Divisor}) \times (\text{Quotient}) + \text{Remainder}$$ where the degree of the remainder is less than the degree of the divisor.
3. **Set up polynomial long division:** Divide the leading term of the dividend $$-8x^4$$ by the leading term of the divisor $$4x$$:
$$\frac{-8x^4}{4x} = -2x^3$$
4. **Multiply divisor by this term:**
$$-2x^3 \times (4x - 3) = -8x^4 + 6x^3$$
5. **Subtract this from the dividend:**
$$(-8x^4 + 22x^3 - 25x + 9) - (-8x^4 + 6x^3) = 22x^3 - 6x^3 - 25x + 9 = 16x^3 - 25x + 9$$
6. **Repeat the process:** Divide the leading term $$16x^3$$ by $$4x$$:
$$\frac{16x^3}{4x} = 4x^2$$
7. **Multiply divisor by $$4x^2$$:**
$$4x^2 \times (4x - 3) = 16x^3 - 12x^2$$
8. **Subtract:**
$$ (16x^3 - 25x + 9) - (16x^3 - 12x^2) = 12x^2 - 25x + 9$$
9. **Divide leading term $$12x^2$$ by $$4x$$:**
$$\frac{12x^2}{4x} = 3x$$
10. **Multiply divisor by $$3x$$:**
$$3x \times (4x - 3) = 12x^2 - 9x$$
11. **Subtract:**
$$ (12x^2 - 25x + 9) - (12x^2 - 9x) = -16x + 9$$
12. **Divide leading term $$-16x$$ by $$4x$$:**
$$\frac{-16x}{4x} = -4$$
13. **Multiply divisor by $$-4$$:**
$$-4 \times (4x - 3) = -16x + 12$$
14. **Subtract:**
$$ (-16x + 9) - (-16x + 12) = 9 - 12 = -3$$
15. **Since the remainder $$-3$$ is a constant and degree less than divisor degree 1, stop here.**
16. **Write the quotient and remainder:**
$$\text{Quotient} = -2x^3 + 4x^2 + 3x - 4$$
$$\text{Remainder} = -3$$
17. **Check if divisor is a factor:** Since remainder $$\neq 0$$, the divisor $$4x - 3$$ is **not** a factor of the dividend.
**Final answers:**
- Quotient: $$-2x^3 + 4x^2 + 3x - 4$$
- Remainder: $$-3$$
- Is $$4x - 3$$ a factor? **False**