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**Hint:**

We are asked to find the value of k such that 3 is zero of the polynomial \[2{{x}^{2}}+x+k.\] We will first learn the definition of zero of a polynomial and then we will consider that x = 3 is a zero and compare P(3) = 0 to find the value of k where \[P\left( x \right)=2{{x}^{2}}+x+k.\]

**Complete step by step answer:**

We are given a polynomial \[2{{x}^{2}}+x+k\] and mentioned that we have to find such a value of k so that 3 is a root of the given above polynomial. To solve for k, we will first learn about the roots (zeroes). Zeroes (roots) are those values of x which when inserted in the place of x in the polynomial P(x) then the result will be zero such that P(x) = 0 are called zeros of the polynomial.

For example if we take P(x) = x + 2 that is at x = – 2. We have,

\[P\left( -2 \right)=-2+2=0\]

So, – 2 is zero of P(x) = x + 2. Similarly, let \[P\left( x \right)={{x}^{2}}-1\] so x = 1 is zero of P(x).

Now, we are having a polynomial \[2{{x}^{2}}+x+k\] we are asked to find k such that 3 is zero of the polynomial.

Let x = 3 is zero of \[P\left( x \right)=2{{x}^{2}}+x+k.\] So by definition of zero, we put x = 3, we get P(3) = 0. So,

\[P\left( 3 \right)=2{{\left( 3 \right)}^{2}}+3+k=0\]

Now, we simplify,

\[\Rightarrow 2\times 9+3+k=0\left[ \text{As }{{3}^{2}}=9 \right]\]

\[\Rightarrow 18+3+k=0\]

So solving, we get,

\[\Rightarrow 21+k=0\]

This implies we have,

\[\Rightarrow k=-21\]

Hence, we got that for k = – 21, x = 3 is a zero of the polynomial, \[2{{x}^{2}}+x+k.\]

We can cross-check as below. We will put k = – 21 in \[2{{x}^{2}}+x+k\] so we have

\[P\left( x \right)=2{{x}^{2}}+x-21\]

Now we will find its zero. So, by splitting the middle term, we have,

\[\Rightarrow 2{{x}^{2}}+x-21=2{{x}^{2}}+7x-6x-21\]

On simplifying, we get,

\[\Rightarrow 2{{x}^{2}}+x-21=x\left( 2x+7 \right)-3\left( 2x+7 \right)\]

On further solving, we get,

\[\Rightarrow 2{{x}^{2}}+x-21=\left( x-3 \right)\left( 2x+7 \right)\]

So, we get,

\[\Rightarrow x-3=0;2x+7-0\]

\[\Rightarrow x=-3;x=\dfrac{-7}{2}\]

So, the zeroes are 3 and \[\dfrac{-7}{2}.\]

Hence, 3 is a zero, so we get k as – 21.

**Note:**

There is an easy way to check this. We have \[2{{x}^{2}}+x-21,\] we have to check if 3 is a zero of this, so we put x = 3 and check we get 0 as an answer or not. So, we get,

\[P\left( 3 \right)=2{{\left( 3 \right)}^{2}}+3-21\]

Simplifying, we get,

\[P\left( 3 \right)=2\times 9+3-21\]

\[\Rightarrow P\left( 3 \right)=18+3-21\]

\[\Rightarrow P\left( 3 \right)=21-21\]

\[\Rightarrow P\left( 3 \right)=0\]

As we get P(3) = 0, so k = – 21 is the correct solution.

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