Question

How do you find the polynomial function with leading coefficient 2 that has a given degree and zeros: degree 3; zeroes $2, \dfrac{1}{2}, \dfrac{3}{2}$?

Answer 1

Hint: This question is from the topic of polynomials. In this question, we will find the polynomial function using the given data. In solving this question, using the given zeros, we will write the polynomial function. After that, we will multiply the leading coefficient 2 with the function. After solving the further question, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find the polynomial function whose leading coefficient is 2. And, it is also said that the polynomial function is of degree 3. The zeroes of polynomial function are given as 2, \[\dfrac{1}{2}\], and \[\dfrac{3}{2}\].
Before solving this question, we should know that the degree of any polynomial function is always equal to the number of zeroes of that polynomial function.
So, the number of zeroes of polynomial function is 3 and they are 2, \[\dfrac{1}{2}\], and \[\dfrac{3}{2}\].
As we know that if ‘a’ is the zero of any polynomial function, then (x-a) will be the root of that polynomial function, so we can say that \[\left( x-2 \right)\], \[\left( x-\dfrac{1}{2} \right)\], and \[\left( x-\dfrac{3}{2} \right)\] are the roots (or we can say the factors) of the polynomial function.
Hence, we can say that the polynomial function will be like:
\[\left( x-2 \right)\left( x-\dfrac{1}{2} \right)\left( x-\dfrac{3}{2} \right)\]
But, it is given that the leading co-efficient is 2. So, we will the polynomial function as
\[2\left( x-2 \right)\left( x-\dfrac{1}{2} \right)\left( x-\dfrac{3}{2} \right)\]
Using foil method formula: \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\], the above can also be written as
\[\Rightarrow 2\left( {{x}^{2}}-\dfrac{x}{2}-2x+\left( -2 \right)\left( -\dfrac{1}{2} \right) \right)\left( x-\dfrac{3}{2} \right)\]
The above can also be written as
\[\Rightarrow 2\left( {{x}^{2}}-\dfrac{5x}{2}+1 \right)\left( x-\dfrac{3}{2} \right)\]
Taking 2 inside the bracket, we get
\[\Rightarrow \left( 2{{x}^{2}}-2\times \dfrac{5x}{2}+2\times 1 \right)\left( x-\dfrac{3}{2} \right)\]
\[\Rightarrow \left( 2{{x}^{2}}-5x+2 \right)\left( x-\dfrac{3}{2} \right)\]
Again applying the foil method, we can write the above as
\[\Rightarrow \left( 2{{x}^{2}} \right)\left( x-\dfrac{3}{2} \right)-\left( 5x \right)\left( x-\dfrac{3}{2} \right)+\left( 2 \right)\left( x-\dfrac{3}{2} \right)\]
Applying it further, we get it as
\[\Rightarrow \left( 2{{x}^{2}} \right)\left( x \right)-\left( 2{{x}^{2}} \right)\left( \dfrac{3}{2} \right)-\left( 5x \right)\left( x \right)-\left( 5x \right)\left( -\dfrac{3}{2} \right)+\left( 2 \right)\left( x \right)-\left( 2 \right)\left( \dfrac{3}{2} \right)\]
The above can also be written as
\[\Rightarrow 2{{x}^{3}}-3{{x}^{2}}-5{{x}^{2}}+\dfrac{15x}{2}+2x-3\]
The above can also be written as
\[\Rightarrow 2{{x}^{3}}-8{{x}^{2}}+\dfrac{19x}{2}-3\]
Hence, we get that the polynomial function with leading coefficient 2 and having zeroes as 2, \[\dfrac{1}{2}\], and \[\dfrac{3}{2}\] is \[2{{x}^{3}}-8{{x}^{2}}+\dfrac{19x}{2}-3\].

Note: We should have a proper knowledge in the topic of polynomials to solve this type of question easily. We should know that the number of zeros of a polynomial function is always equal to the degree of that polynomial function.
Remember the foil method formula. The formula is: \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\]