Question

In expansion of (2x+1)(x^2+px+4), the coefficient of x is twice the coefficient of x^2.

Find the value of x​

Answer 1

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`

We're told that

`8+p=2(1+2p)`

which we can use to solve for p:

`8+p=2+4p`

`6=3p`

`\boxed{p=2}`

Answer 2

To find the value of x in the given expression, we expand it and equate the coefficients of x and x^2.

Step-by-step explanation:

To find the value of x in the expansion of (2x+1)(x^2+px+4), where the coefficient of x is twice the coefficient of x^2, we need to expand the expression and equate the coefficients.

After expanding the expression, we get 2x^3 + (2p+1)x^2 + (8+p)x+4.

Since the coefficient of x is twice the coefficient of x^2, we can equate the coefficients: 2p+1 = 2(8+p). Solving this equation will give us the value of x.

Learn more about Expanding expressions