Find P Value A28211

1. **State the problem:** We need to find the value of $p$ such that in the expansion of $px^2(5 + px)^8$, the coefficient of the $x^6$ term is $3402$. 2. **Understand the expression:** The expression is $px^2(5 + px)^8$. When expanded, the powers of $x$ come from both $x^2$ and $(5 + px)^8$. 3. **General term of the binomial expansion:** The general term of $(5 + px)^8$ is given by $$T_{k+1} = \binom{8}{k} 5^{8-k} (px)^k = \binom{8}{k} 5^{8-k} p^k x^k$$ where $k = 0, 1, 2, ..., 8$. 4. **Combine with $px^2$:** Multiplying by $px^2$ gives $$px^2 \times T_{k+1} = p x^2 \times \binom{8}{k} 5^{8-k} p^k x^k = \binom{8}{k} 5^{8-k} p^{k+1} x^{k+2}$$ 5. **Find the term with $x^6$:** We want the power of $x$ to be $6$, so $$k + 2 = 6 \implies k = 4$$ 6. **Coefficient of $x^6$ term:** Substitute $k=4$: $$\text{Coefficient} = \binom{8}{4} 5^{8-4} p^{4+1} = \binom{8}{4} 5^4 p^5$$ 7. **Calculate binomial coefficient and powers:** $$\binom{8}{4} = \frac{8!}{4!4!} = 70$$ $$5^4 = 625$$ 8. **Set coefficient equal to 3402:** $$70 \times 625 \times p^5 = 3402$$ 9. **Simplify:** $$43750 p^5 = 3402$$ 10. **Solve for $p^5$:** $$p^5 = \frac{3402}{43750} = 0.07774$$ 11. **Find $p$:** $$p = \sqrt[5]{0.07774} \approx 0.63$$ **Final answer:** $$p \approx 0.63$$