You're probably supposed to find the value of p and not x, which is impossible to do.
Expanding gives
![(2x+1)(x^2+px+4)=2x^3+(1+2p)x^2+(8+p)x+4](https://tex.z-dn.net/?f=%282x%2B1%29%28x%5E2%2Bpx%2B4%29%3D2x%5E3%2B%281%2B2p%29x%5E2%2B%288%2Bp%29x%2B4)
We're told that
![8+p=2(1+2p)](https://tex.z-dn.net/?f=8%2Bp%3D2%281%2B2p%29)
which we can use to solve for p:
![8+p=2+4p](https://tex.z-dn.net/?f=8%2Bp%3D2%2B4p)
![6=3p](https://tex.z-dn.net/?f=6%3D3p)
![\boxed{p=2}](https://tex.z-dn.net/?f=%5Cboxed%7Bp%3D2%7D)
Answer:
50 - 2c
Step-by-step explanation:
i think its incomplete
The answer is 7 because add them to see the distance since it’s both positive if you count the distance it is away from zero so 3+4=7
Given recursive formula is
f(n) = 22 * f(n - 1)
Where first term : f(1) = 10
And we need to find the first five terms of this sequence. So , let's find the second term by plug in n = 2 in the above formula. Therefore,
f (2) = 2.2 * f ( 2 -1)
= 2.2 * f(1)
= 2.2 * 10
= 22
So second term is 22.
Similarly plug in 3 for n to get the third term. Hence,
f(3) = 2.2 * f(3 -1) = 2.2 * f(2) = 2.2 * 22= 48.4
f(4) = 2.2 *f( 4 -1) = 2.2 * f(3) = 2.2* 48.4 = 106.48
f(5) = 2.2* f(5-1)= 2.2 * f(4) = 2.2 * 106.48 = 234.256
So, the first five terms are 10, 22, 48.4, 106.48 and 234.256.
Hope this helps you!