Remainder W Value 030612

1. **State the problem:** We want to find the value of $w$ such that when the polynomial $$f(x) = 5x^4 + 9x^3 - 7x^2 + wx - 3$$ is divided by $x + 2$, the remainder is 3. 2. **Recall the Remainder Theorem:** When a polynomial $f(x)$ is divided by $x - a$, the remainder is $f(a)$. Here, the divisor is $x + 2$, which can be rewritten as $x - (-2)$, so $a = -2$. 3. **Apply the Remainder Theorem:** The remainder when dividing by $x + 2$ is $f(-2)$. We want: $$f(-2) = 3$$ 4. **Calculate $f(-2)$:** $$f(-2) = 5(-2)^4 + 9(-2)^3 - 7(-2)^2 + w(-2) - 3$$ Calculate powers: $$(-2)^4 = 16$$ $$(-2)^3 = -8$$ $$(-2)^2 = 4$$ Substitute: $$f(-2) = 5 \times 16 + 9 \times (-8) - 7 \times 4 + w \times (-2) - 3$$ Simplify: $$= 80 - 72 - 28 - 2w - 3$$ Combine constants: $$80 - 72 = 8$$ $$8 - 28 = -20$$ $$-20 - 3 = -23$$ So: $$f(-2) = -23 - 2w$$ 5. **Set remainder equal to 3:** $$-23 - 2w = 3$$ 6. **Solve for $w$:** Add 23 to both sides: $$-2w = 3 + 23$$ $$-2w = 26$$ Divide both sides by $\cancel{-2}$: $$w = \frac{26}{\cancel{-2}} \times \cancel{-1} = -13$$ 7. **Final answer:** $$\boxed{w = -13}$$