Question

Find the value of $ p,\,2{x^3} + p{x^2} + 11x + p + 3 $ is exactly divisible by $ 2x - 1 $ .

Answer 1

Hint: We will find the value of $ x $ into the polynomial by equating it to zero.. Therefore, we will put the value into the given polynomial to calculate the value of $ P $ .

Complete step-by-step answer:
Let $ p(x) = 2{x^3} + P{x^2} + 11x + 1 + 3 $
And $ q(x) = 2x - 1 $
If $ p(x) $ is exactly divisible by $ q(x) $ then $ q(x) $ is a factor of $ p(x) $ .
Then $ q(x) = 0 $
 $
   \Rightarrow 2x - 1 = 0 \\
   \Rightarrow 2x = 0 + 1 \\
   \Rightarrow 2x = 1 \\
   \Rightarrow x = \dfrac{1}{2} \\
  $
Now, we will substitute the value of $ x = \dfrac{1}{2} $ in the $ p(x) $ polynomial, we have
 $ p\left( {\dfrac{1}{2}} \right) = 2 \times {\left( {\dfrac{1}{2}} \right)^3} + p{\left( {\dfrac{1}{2}} \right)^2} + 11\left( {\dfrac{1}{2}} \right) + p + 3 $
 $ p\left( {\dfrac{1}{2}} \right) = 2 \times \dfrac{1}{8} + p \times \dfrac{1}{4} + \dfrac{{11}}{2} + p + 3 $
 $ 0 = \dfrac{1}{4} + \dfrac{p}{4} + \dfrac{{11}}{2} + p + 3 $
We will take 4 LCM of 4,4,2,we have
 $ \Rightarrow \dfrac{{1 + p + 22 + 4p + 12}}{4} = 0 $
 $ \Rightarrow \dfrac{{35 + 4p + p}}{4} = 0 $
 $ \Rightarrow \dfrac{{35 + 5p}}{4} = 0 $
 $ \Rightarrow 35 + 5p = 0 $
 $ \Rightarrow 5p = - 35 $
 $ \Rightarrow p = \dfrac{{ - 35}}{5} $
 $ \Rightarrow p = - 7 $
Hence, the value of $ p = - 7 $

Note: Students must know that if $ q(x) $ is a factor of $ p(x) $ , then remainder is always $ 0 $ . We will use this approach to solve the question. This knowledge can be applied to polynomials as well. When two or more polynomials are multiplied, we call each of these polynomials factors of the product. ... Also, one polynomial is divisible by another polynomial if the quotient is also a polynomial.