Question

How do you write a polynomial function of least degree with an integral coefficient that has the given zeros $ -3,\dfrac{-1}{3},5 $?

Answer 1

Hint: In this question, we are given three zeros of some polynomial f(x). We need to determine this f(x) where coefficients are all integral. We will use the factor theorem. According to the factor theorem, if n is a zero of the equation f(x) = 0 then (x-n) is a factor of f(x). Using this we will multiply all three factors and try to form an equation with the integral coefficient.
 Here we are given the roots of the equation as $ -3,\dfrac{-1}{3},5 $ . We need to find the equation with the integral coefficient. Let the polynomial function by f(x). Let us use the factor theorem here. According to the factor theorem, if n is a zero of the equation f(x) then (x-n) is a factor of f(x). So here we have zeroes of the f(x) as $ -3,\dfrac{-1}{3},5 $ .
So the factor of f(x) will be $ \left( x-\left( -3 \right) \right),\left( x-\left( \dfrac{-1}{3} \right) \right),\left( x-5 \right) $ i.e. factors of f(x) are $ \left( x+3 \right),\left( x+\dfrac{1}{3} \right),\left( x-5 \right) $ .

Complete step by step answer:
We know that, multiplying all the factors of a polynomial gives us the original polynomial so we can say $ f\left( x \right)=\left( x+3 \right),\left( x+\dfrac{1}{3} \right),\left( x-5 \right) $ .
Let us look at the factor $ \left( x+\dfrac{1}{3} \right) $ . We need integral coefficient for polynomial but multiplying by $ \dfrac{1}{3} $ will not give us integral coefficient as $ \dfrac{1}{3} $ is not integer, so we need to change it.
 $ \left( x+\dfrac{1}{3} \right) $ can also be written as $ \left( \dfrac{3x+1}{3} \right) $ . When we will put this factor as equal to zero $ \left( \dfrac{3x+1}{3} \right)=0 $ 3 will have no use, so we can write factor (3x+1) instead of $ \left( x+\dfrac{1}{3} \right) $ .
3x+1 = 0 will give us $ x=-\dfrac{1}{3} $ . So we can use it.
Therefore, our polynomial function becomes, $ f\left( x \right)=\left( x+3 \right),\left( 3x+1 \right),\left( x-5 \right) $ .
Let us multiply these factors, multiplying (x+3) by (3x+1) we get,
 $ \begin{align}
  & f\left( x \right)=\left( x\cdot 3x+3\cdot 3x+x\cdot 1+3\cdot 1 \right)\left( x-5 \right) \\
 & \Rightarrow f\left( x \right)=\left( 3{{x}^{2}}+9x+x+3 \right)\left( x-5 \right) \\
 & \Rightarrow f\left( x \right)=\left( 3{{x}^{2}}+10x+3 \right)\left( x-5 \right) \\
\end{align} $
Now multiplying these two factors we get,
 $ f\left( x \right)=3{{x}^{2}}\cdot x+10x\cdot x+3x+3{{x}^{2}}\cdot \left( -5 \right)+10x\cdot \left( -5 \right)+3\cdot \left( -5 \right) $ .
Simplifying the terms we get,
 $ f\left( x \right)=3{{x}^{3}}+10{{x}^{2}}+3x-15{{x}^{2}}-50x-15 $ .
Now let us add and subtract the like terms we get,
 $ f\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-47x-15 $ .
As we can see all the coefficients (3, -5, -47, -15) are integers and it cannot be reduced so this is our required polynomial function.
$ 3{{x}^{3}}-5{{x}^{2}}-47x-15 $ is the required answer.

Note:
 Students should always check that while transforming a factor, the new factor should also give the same zero as given by the previous factor $ \left( x+\dfrac{1}{3}\text{ and }3x+1 \right) $ . Take care of signs while multiplying two polynomials. Note that, for n to be zero (x-n) is a factor and not (x+n).