Question

How do you write a polynomial with zeros : \[ - 2\], multiplicity: 2; 4, multiplicity: 1; degree: 3?

Answer 1

Hint: Here, we need to write the polynomial that has the given zeros and degree. We will use the factor theorem to find the factors of the polynomial. Then, we will multiply the factors using the distributive law of multiplication and the FOIL method to find the required polynomial with degree 3.

Formula used:
 The factor theorem states that if \[x = k\] is a zero of a polynomial \[p\left( x \right)\], then \[x - k\] is a factor of the polynomial \[p\left( x \right)\].

Complete step-by-step answer:
The required polynomial has degree 3.
This means that the highest power of the variable in the polynomial is 3.
Thus, the polynomial is a cubic polynomial.
A cubic polynomial always has 3 zeroes.
It is given that \[ - 2\] is a zero of the polynomial with multiplicity 2.
This means that two of the three zeroes of the required polynomial are \[ - 2\].
It is given that 4 is a zero of the polynomial with multiplicity 1.
This means that last of the three zeroes of the required polynomial is 4.
Thus, the zeroes of the polynomial are \[ - 2\], \[ - 2\], and 4.
Now, we will use the factor theorem.
The factor theorem states that if \[x = k\] is a zero of a polynomial \[p\left( x \right)\], then \[x - k\] is a factor of the polynomial \[p\left( x \right)\].
Therefore, we get the three factors of the required polynomial as \[x - \left( { - 2} \right)\], \[x - \left( { - 2} \right)\], and \[x - 4\].
The required polynomial is the product of its three factors.
Let the required polynomial be \[p\left( x \right)\].
Therefore, we get
\[p\left( x \right) = \left[ {x - \left( { - 2} \right)} \right]\left[ {x - \left( { - 2} \right)} \right]\left[ {x - 4} \right]\]
Simplifying the expression, we get
\[p\left( x \right) = \left( {x + 2} \right)\left( {x + 2} \right)\left( {x - 4} \right)\]
The FOIL method stands for First, Outer, Inner, Last.
First means multiplication of the first term of the first parentheses and the first term of the second parentheses.
Outer means multiplication of the first term of the first parentheses and the second term of the second parentheses.
Inner means multiplication of the second term of the first parentheses and the first term of the second parentheses.
Last means multiplication of the second term of the first parentheses and the second term of the second parentheses.
The FOIL method can be represented by the equation \[\left( {a + b} \right)\left( {c + d} \right) = a \cdot c + a \cdot d + b \cdot c + b \cdot d\].
Multiplying the first and second parentheses using the FOIL method, we get
\[ \Rightarrow p\left( x \right) = \left( {{x^2} + 2x + 2x + 4} \right)\left( {x - 4} \right)\]
Adding the like terms, we get
\[ \Rightarrow p\left( x \right) = \left( {{x^2} + 4x + 4} \right)\left( {x - 4} \right)\]
Now, the right hand side is the product of a trinomial and a binomial.
We will group two terms in the trinomial.
Rewriting the expression, we get
\[ \Rightarrow p\left( x \right) = \left[ {{x^2} + \left( {4x + 4} \right)} \right]\left( {x - 4} \right)\]
Multiplying the two parentheses with the help of FOIL method, we get
\[ \Rightarrow p\left( x \right) = {x^2} \cdot x + {x^2}\left( { - 4} \right) + \left( {4x + 4} \right)x + \left( {4x + 4} \right)\left( { - 4} \right)\]
Simplifying the terms in the expression, we get
\[ \Rightarrow p\left( x \right) = {x^3} - 4{x^2} + x\left( {4x + 4} \right) - 4\left( {4x + 4} \right)\]
Now, we can simplify the remaining expression using the distributive law of multiplication.
Multiplying the terms using the distributive law of multiplication \[a\left( {b + c} \right) = a \cdot b + a \cdot c\], we get
\[ \Rightarrow p\left( x \right) = {x^3} - 4{x^2} + 4{x^2} + 4x - 16x - 16\]
Combining the like terms in the expression, we get
\[ \Rightarrow p\left( x \right) = {x^3} - 12x - 16\]
Therefore, we get the required polynomial as \[{x^3} - 12x - 16\].

Note: We have used the distributive law of multiplication in the solution to multiply a monomial by a binomial in the solution. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
Verification: We can verify our polynomial by putting the given zeros and checking if the value of the polynomial becomes 0.
Substituting \[x = - 2\] in the polynomial \[p\left( x \right) = {x^3} - 12x - 16\], we get
\[ \Rightarrow p\left( { - 2} \right) = {\left( { - 2} \right)^3} - 12\left( { - 2} \right) - 16\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow p\left( { - 2} \right) = - 8 + 24 - 16\\ \Rightarrow p\left( { - 2} \right) = 0\end{array}\]
Substituting \[x = 4\] in the polynomial \[p\left( x \right) = {x^3} - 12x - 16\], we get
\[ \Rightarrow p\left( 4 \right) = {4^3} - 12\left( 4 \right) - 16\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow p\left( 4 \right) = 64 - 48 - 16\\ \Rightarrow p\left( 4 \right) = 0\end{array}\]
Since the value of the polynomial becomes 0 at \[x = - 2\] and \[x = 4\], we have verified that \[ - 2\] and 4 are the zeroes of the polynomial \[p\left( x \right) = {x^3} - 12x - 16\], and our polynomial is correct.