Answer:
a) 2x^4 + 8x^3 + x^2 + 3x
b) evaluate this polynomial at x = 10
Explanation:
Hi!
a)
The product of a sum is the sum of the products, having said this:
![(2x^3 + 8x^2 + x + 3) * x = 2x^4 + 8x^3 + x^2 + 3x](https://img.qammunity.org/2020/formulas/mathematics/high-school/hlutrojpmktkelpvbuddfearwbuzkkp2wp.png)
b)
We can use the result from part a and consider x = 10.
First lets take the first polynomial
![2x^3 + 8x^2 + x + 3 = 2(10)^3 + 8 (10)^2+10^1+3(10)^0](https://img.qammunity.org/2020/formulas/mathematics/high-school/5dulgczjo22oqx2naiuo5elgyn9ysfik0d.png)
Knowing that every number (excepting probably 0, which could be undefined for some or 1 for others) to the power of 0 is one:
![2x^3 + 8x^2 + x + 3 = 2(1000)+ 8(100)+10+3(1) = 2813](https://img.qammunity.org/2020/formulas/mathematics/high-school/xhpj73r3lopznc4fys5rlfvydqhg34kztn.png)
Which is the initial number
Now lets conisder the product of the polynomial times x and evaluate it at x=10:
![(2x^3 + 8x^2 + x + 3) * x = 2x^4 + 8x^3 + x^2 + 3x](https://img.qammunity.org/2020/formulas/mathematics/high-school/hlutrojpmktkelpvbuddfearwbuzkkp2wp.png)
![(2(10)^4 + 8(10)^3 + 10^2 + 3(10) = 2 (10000) + 8(1000) + 100 + 3(10)](https://img.qammunity.org/2020/formulas/mathematics/high-school/q2pr5vtmgsaeh1mi28nnukzefu7xl8283u.png)
![= 28130](https://img.qammunity.org/2020/formulas/mathematics/high-school/c8shr3ugpebulup3itzflya48of8v534tr.png)
We can now see that teh rule is treu because we can see the whole number as a sum of numbers from 0 to 9 multiplied by a power of 10, and when it is multiplied by 10, each power increases by one, and at the end, the final result is adding a zero at the end