Question

Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients-
$f(x) = 2x^3 + x^2 - 5x + 2; \dfrac{1}2{}, 1, -2$

Answer 1

Hint: To solve this problem, one should have theoretical knowledge of cubic equations and the relationship between their coefficients and roots. We will substitute the three roots and check if they are equal to zero. Then we will verify the relations using some formulas. The formulas that will be used to verify this relationship are-
$\mathrm\alpha+\mathrm\beta+\mathrm\gamma=\dfrac{-\mathrm b}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{\mathrm\beta\mathrm\gamma}+\mathrm{\mathrm\gamma\mathrm\alpha}=\dfrac{\mathrm c}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta\mathrm\gamma}=\dfrac{-\mathrm d}{\mathrm a}$

Complete step by step answer:
First we will verify that the given numbers are roots of f(x)-
$\mathrm f\left(\dfrac12\right)=2\left(\dfrac12\right)^3+\left(\dfrac12\right)^2-5\left(\dfrac12\right)+2\\\mathrm f\left(\dfrac12\right)=\dfrac28+\dfrac14-\dfrac52+2\\\mathrm f\left(\dfrac12\right)=-\dfrac42+2=0$
Hence $\dfrac{1}{2}$ is a root of f(x).
$f(1) = 2(1)^3 + (1)^2 - 5(1) + 2$
$f(1) = 2 + 1 - 5 + 2 = 0$
Hence 1 is a root of f(x).
$f(-2) = 2(-2)^3 + (-2)^2 -5(-2) + 2$
$f(-2) = 2(-8) + 4 + 10 +2$
$f(-2) = -16 +4 +10 +2 = 0$
Hence -2 is a root of f(x).
Now, we have to verify the relationship between the roots and coefficients using the given formulas-
In the given cubic equation-
$a = 2, b = 1, c = -5, d = 2$
$\begin{align}
  &{{\alpha }} + {{\beta }} + {{\gamma }} = \dfrac{1}{2} + 1 - 2 = - \dfrac{1}{2} \\
   &- \dfrac{{\text{b}}}{{\text{a}}} = - \dfrac{1}{2} \\
  &Hence,\;{{\alpha }} + {{\beta }} + {{\gamma }} = - \dfrac{{\text{b}}}{{\text{a}}} \\
  &\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{1}{2} \times 1 - 1 \times 2 - 2 \times \dfrac{1}{2} = \dfrac{1}{2} - 2 - 1 = - \dfrac{5}{2} \\
&\dfrac{{\text{c}}}{{\text{a}}} = - \dfrac{5}{2} \\
  &Hence,\;\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{{\text{c}}}{{\text{a}}} \\
  &\alpha \beta \gamma = \dfrac{1}{2} \times 1 \times \left( { - 2} \right) = - 1 \\
   &- \dfrac{{\text{d}}}{{\text{a}}} = - \dfrac{2}{2} = - 1 \\
  &Hence,\;\alpha \beta \gamma = \dfrac{{\text{d}}}{{\text{a}}} \\
\end{align} $
All the given conditions are verified.

Note:
The above three conditions signify that the sum of roots is equal to the negative of ratio of coefficients of $x^2$ and $x^3$. Also, the sum of the product of roots taken two at a time is equal to the ratio of coefficients of $x$ and $x^3$. The product of roots is equal to the ratio of constant and the coefficient of $x^3$.