Question

If the zeros of the polynomial\[p(x) = {x^3} - 3{x^2} + x + 1\]are $ a - b,a,a + b $
Find $ a $ and $ b $ .

Answer 1

Hint: First, assume that a cubic equation in any form and zeros of the polynomial is given.
Cubic means at most power three terms, applied by the condition that the sum of the given roots and product of the given roots.
After finding the values of the sum of the two terms and the product of the two terms, we can easily find the values of $ a $ and $ b $
Formula used: The general cubic equation formula is $ p(x) = a{x^3} + b{x^2} + cx + d $
The sum of the given zeros can be expressed as $ \dfrac{{ - b}}{a} $ and the sum of product zeroes is $ \dfrac{c}{a} $

Complete step by step answer:
To solve the equation, let us fix $ \alpha = a - b,\beta = a,\gamma = a + b $ , also from the general equation we can say that the sum of the zeroes is expressed as $ \alpha + \beta + \gamma = \dfrac{{ - b}}{a} $ and the sum of the product zeroes are $ \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} $
Now compare the general equation with the given equation, that is $ a{x^3} + b{x^2} + cx + d = {x^3} - 3{x^2} + x + 1 $ from this we get $ a = 1,b = - 3,c = 1,d = 1 $
Apply these values in the sum of the zeroes formula, we get $ \alpha + \beta + \gamma = \dfrac{{ - b}}{a} \Rightarrow (a - b) + a + (a + b) = \dfrac{{ - ( - 3)}}{1} $
Thus, further solving we get, $ (a) + a + (a) = 3 $ (where common values in opposite signs get canceled)
Hence, we get the sum of the zeroes as $ (a) + a + (a) = 3 \Rightarrow 3a = 3 \Rightarrow a = 1 $
Now, to find the sum of the product zeroes, we have $ \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} $ and apply the values in above we get; $ \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} \Rightarrow (a - b)a + a(a + b) + (a + b)(a - b) = \dfrac{1}{1} $
Further solving the equation, we get, $ (a - b)a + a(a + b) + (a + b)(a - b) = 1 \Rightarrow {a^2} - ba + {a^2} + ab + {a^2} - {b^2} = 1 $
Canceling the common terms, we get $ {a^2} + {a^2} + {a^2} - {b^2} = 1 \Rightarrow 3{a^2} - {b^2} = 1 $
Now substituting the value of $ a = 1 $ from the sum of the zeroes, then we get $ 3{a^2} - {b^2} = 1 \Rightarrow 3{(1)^2} - {b^2} = 1 \Rightarrow - {b^2} = 1 - 3 $
Further solving the equation, we get $ - {b^2} = 1 - 3 \Rightarrow {b^2} = 2 \Rightarrow b = \pm \sqrt 2 $ (after taking the square root)
Hence, we get the value of $ a = 1 $ and $ b = \pm \sqrt 2 $

Note: If the value $ a = 0 $ then the function is changed as a quadratic function because $ p(x) = a{x^3} + b{x^2} + cx + d \Rightarrow b{x^2} + cx + d $ . So that value of a will never be zero.
The addition is the sum of adding two or more than two numbers as variables.
We will see what multiplication is, Multiplicand refers to the number multiplied. The multiplier is the number that refers to the number which multiplies the first number.
Be careful with the sum of the zeroes, the formula is $ \dfrac{{ - b}}{a} $ and not $ \dfrac{b}{a} $