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For what value of k is 3 a zero of the polynomial \[2{{x}^{2}}+x+k.\]

Last updated date: 13th Jun 2024
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 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.

 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.