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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​

User Andzep
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2 Answers

5 votes

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}

User Dave Huang
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8.3k points
3 votes

Final answer:

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

User Bill Goldberg
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