Verified Answer :

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ANSWER

\(p(x)=-3+6(x+2)^2\)


EXPLANATION


The given function is


\(p(x)=21+24x+6x^2\)


We make the coefficient of \(x^2\) unity by factoring \(6\) out of the last two terms to obtain;


\(p(x)=21+6(4x+x^2)\)


We now add and subtract half the coefficient of \(x\) multiplied by a factor of 6 to obtain;


\(p(x)=21-6(2)^2+6(2)^2+6(4x+x^2)\)


We now factor 6 out of the last two terms to get;


\(p(x)=21-6\times 4+6((2)^2+4x+x^2)\)


\(p(x)=21-24+6((2)^2+4x+x^2)\)


The quadratic trinomial in the parenthesis is now a perfect square.


\(p(x)=-3+6(x+2)^2\)


Hence the vertex form of the polynomial is


\(p(x)=6(x+2)^2-3\)









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The vertex form of p(x) will be \(p(x)=6(x+2)^2-3\).

Given function,

\(p(x) = 21 + 24x + 6x^2\).

We have to write p(x) in vertex form.

Now,

\(p(x) = 6x^2+ 24x + 21\)

\(p(x)=6(x^{2} +4x)+21\)

How to write the vertex form?

We now add and subtract half the coefficient of x multiplied by a factor of 6, we get

\(p(x)=6(x^{2} +4x)+2^{2} \times 6-2^{2} \times 6+21\)

\(p(x)=6(x^{2} +4x+2^{2} )-24+21\\\)

\(p(x)=6(x+2)^2-3\)

Since the vertex form of polynomial compulsorily contains the perfect square of x.

So the vertex form of polynomial p(x) will be

\(p(x)=6(x+2)^2-3\).

Hence the vertex form of p(x) will be \(p(x)=6(x+2)^2-3\).

For more details about vertex form, follow the link:

https://brainly.com/question/8609063

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