Question

The polynomial $ p{x^3} + 4{x^2} - 3x + q $ is completely divisible by $ {x^2} - 1 $ ; find the values of p and q. Also, for these values of p and q factorize the given polynomial completely.

Answer 1

Hint: The given polynomial $ p{x^3} + 4{x^2} - 3x + q $ is completely divisible by its divisor $ {x^2} - 1 $ , this means the remainder is zero. As we can see the divisor is a quadratic polynomial, find its factors. These factors will also be the factors of the dividend, which means they are the roots. If $ \left( {x - a} \right) $ is a root of a function $ f\left( x \right) $ , then the value of $ f\left( a \right) $ is equal to zero. Use this info to solve for p and q. And after finding the values of p and q, divide the dividend with the divisor to get the quotient.

Complete step-by-step answer:
We are given a polynomial $ p{x^3} + 4{x^2} - 3x + q $ is completely divisible by $ {x^2} - 1 $
We have to find the values of p and q and the quotient which will be obtained when the division takes place.
Let $ f\left( x \right) = p{x^3} + 4{x^2} - 3x + q $
Given that $ {x^2} - 1 $ is a divisor
 $ {x^2} - 1 $ can also be written as $ \left( {x + 1} \right)\left( {x - 1} \right) $
This means, $ \left( {x + 1} \right) $ and $ \left( {x - 1} \right) $ are also the factors of the polynomial $ f\left( x \right) $
Therefore, $ f\left( 1 \right) = 0,f\left( { - 1} \right) = 0 $
On substituting the value of x as 1, -1, the function $ f\left( x \right) $ will become
 $
  x = 1 \\
   \Rightarrow f\left( 1 \right) = p{\left( 1 \right)^3} + 4{\left( 1 \right)^2} - 3\left( 1 \right) + q = 0 \\
   \Rightarrow p + 4 - 3 + q = 0 \\
   \Rightarrow p + q + 1 = 0 \Rightarrow eq\left( 1 \right) \\
  x = - 1 \\
   \Rightarrow f\left( { - 1} \right) = p{\left( { - 1} \right)^3} + 4{\left( { - 1} \right)^2} - 3\left( { - 1} \right) + q = 0 \\
   \Rightarrow - p + 4 + 3 + q = 0 \\
   \Rightarrow - p + q + 7 = 0 \\
   \Rightarrow p = q + 7 \Rightarrow eq\left( 2 \right) \\
  $
On substituting equation 2 in equation 1, we get
 $
  q + 7 + q + 1 = 0 \\
   \Rightarrow 2q + 8 = 0 \\
   \Rightarrow 2q = - 8 \\
  \therefore q = \dfrac{{ - 8}}{2} = - 4 \\
  $
On substituting the value of q in equation 2, we get
 $
  p = q + 7 \\
   \Rightarrow p = - 4 + 7 = 3 \\
  $
On substituting the obtained values of p and q in the given polynomial, we get
 $ p{x^3} + 4{x^2} - 3x + q = 3{x^3} + 4{x^2} - 3x + \left( { - 4} \right) = 3{x^3} + 4{x^2} - 3x - 4 $
Now divide the above polynomial with $ {x^2} - 1 $
 $
  \dfrac{{3{x^3} + 4{x^2} - 3x - 4}}{{{x^2} - 1}} \\
   = \dfrac{{3{x^3} - 3x + 4{x^2} - 4}}{{{x^2} - 1}} \\
   = \dfrac{{3x\left( {{x^2} - 1} \right) + 4\left( {{x^2} - 1} \right)}}{{{x^2} - 1}} \\
   = \dfrac{{\left( {3x + 4} \right)\left( {{x^2} - 1} \right)}}{{{x^2} - 1}} \\
   = 3x + 4 \\
  $
Therefore, $ 3x + 4 $ is the quotient when $ 3{x^3} + 4{x^2} - 3x - 4 $ is divided by $ {x^2} - 1 $

Note: Degree of a polynomial is the highest of the degrees of the polynomial’s monomials which has non zero coefficients. If the degree of a polynomial is 2, then it will have 2 roots and if the degree of the polynomial is 3, then it will have 3 roots. So, therefore the no. of roots of a polynomial or an equation depends upon its degree. Here, the degree of the given polynomial is 3, so it has 3 roots (factors).