Question

Verify that the given numbers alongside the cubic polynomials are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
(i) $ 2{x^3} + {x^2} - 5x + 2;\dfrac{1}{2},1, - 2 $
(ii) $ {x^3} - 4{x^2} + 5x - 2;\,2,1,1 $

Answer 1

Hint: First we will reduce the equation further if possible. Then we will try to factorise the terms in the equation. Then solve the equation by using the quadratic formula and finally evaluate the value of the variable accordingly.

Complete step by step solution:
We will start off by considering the given equation as a function of p such that $ p(x) = 2{x^3} + {x^2} - 5x + 2 $ .
We know that $ \dfrac{1}{2},1 $ and $ - 2 $ are the roots of the expression then we can say that,
\[\left( {x - \dfrac{1}{2}} \right)\left( {x - 1} \right)\left( {x + 2} \right)\] are the factors of $ p(x) $ .
Since\[\left( {x - \dfrac{1}{2}} \right)\left( {x - 1} \right)\left( {x + 2} \right)\] are the factors of $ p(x) $ we can say that if we substitute the values $ \dfrac{1}{2},1 $ and $ - 2 $ in the equation $ 2{x^3} + {x^2} - 5x + 2 $ we will get the value as zero.
Now we will start substituting the values and check if we get zero.
 $
   = 2{x^3} + {x^2} - 5x + 2 \\
   = 2{\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^2} - 5\left( {\dfrac{1}{2}} \right) + 2 \\
   = 2\left( {\dfrac{1}{8}} \right) + \left( {\dfrac{1}{4}} \right) - \left( {\dfrac{5}{2}} \right) + 2 \\
   = \left( {\dfrac{1}{2}} \right) - \left( {\dfrac{5}{2}} \right) + 2 \\
   = - \left( {\dfrac{4}{2}} \right) + 2 \\
   = - 2 + 2 \\
   = 0 \;
  $
Therefore, $ \left( {\dfrac{1}{2}} \right) $ is a root of the expression $ 2{x^3} + {x^2} - 5x + 2 $ .
 Now we will substitute the value $ 1 $ in the expression.
 $
   = 2{x^3} + {x^2} - 5x + 2 \\
   = 2{\left( 1 \right)^3} + {\left( 1 \right)^2} - 5\left( 1 \right) + 2 \\
   = 2 + 1 - 5 + 2 \\
   = 5 - 5 \\
   = 0 \;
  $
 Therefore, $ 1 $ is a root of the expression $ 2{x^3} + {x^2} - 5x + 2 $ .
Now we will substitute the value $ - 2 $ in the expression.
 $
   = 2{x^3} + {x^2} - 5x + 2 \\
   = 2{\left( { - 2} \right)^3} + {\left( { - 2} \right)^2} - 5\left( { - 2} \right) + 2 \\
   = 2( - 8) + (4) + (10) + 2 \\
   = - 16 + 4 + 10 + 2 \\
   = - 16 - 16 \\
   = 0 \;
  $
Therefore, $ - 2 $ is a root of the expression $ 2{x^3} + {x^2} - 5x + 2 $ .
II.
We will start off by considering the given equation as a function of p such that $ p(x) = {x^3} - 4{x^2} + 5x - 2 $ .
We know that $ 2,1 $ and $ 1 $ are the roots of the expression then we can say that,
\[\left( {x - 2} \right)\left( {x - 1} \right)\] are the factors of $ p(x) $ .
Since\[\left( {x - 2} \right)\left( {x - 1} \right)\] are the factors of $ p(x) $ we can say that if we substitute the values $ 2,1 $ and $ 1 $ in the equation $ {x^3} - 4{x^2} + 5x - 2 $ we will get the value as zero.
Now we will start substituting the values and check if we get zero.
 $
   = {x^3} - 4{x^2} + 5x - 2 \\
   = {(2)^3} - 4{(2)^2} + 5(2) - 2 \\
   = 8 - 4(4) + (10) - 2 \\
   = 8 - 16 + 10 - 2 \\
   = 18 - 18 \\
   = 0 \;
  $
Therefore, $ 2 $ is a root of the expression $ {x^3} - 4{x^2} + 5x - 2 $ .
 Now we will substitute the value $ 1 $ in the expression.
 $
   = {x^3} - 4{x^2} + 5x - 2 \\
   = {(1)^3} - 4{(1)^2} + 5(1) - 2 \\
   = 1 - 4(1) + 5 - 2 \\
   = 1 - 4 + 5 - 2 \\
   = 6 - 6 \\
   = 0 \;
  $
 Therefore, $ 1 $ is a root of the expression $ {x^3} - 4{x^2} + 5x - 2 $ .

Note: Do not solve all the equations simultaneously. Solve all the equations separately, so that you don’t miss any term of the solution. Check if the solution satisfies the original equation completely. If any term of the solution doesn’t satisfy the equation, then that term will not be considered as a part of the solution.