If the zeroes of the polynomial x
^3 – 3x^2+ x + 1 are a – b, a, a + b, then find the value

of a and b.​

1 thought on “If the zeroes of the polynomial x<br /> ^3 – 3x^2+ x + 1 are a – b, a, a + b, then find the value <br /><br />of a and b.​”

1. Answer:

   Answer:

   a = 1

   b = ±√2

   Step-by-step explanation:

   \begin{gathered}p(x) = x^3 – 3x^2 + x + 1\\\\\text{A standard cubic equation is of the form $ax^3 + bx^2 + cx + d$}\\\\\text{On comparing the two equations, we get}\\\\a = 1\\\\b = -3\\\\c= 1\\\\\\d=1\\\\\text{Given zeroes are:}\\\\\alpha = a-b\\\\\beta = a\\\\\gamma = a + b\\\\\text{We know that in a cubic polynomial,}\\\\\text{sum of zeroes = $-\dfrac{b}{a}$}\\\\\\\implies \alpha + \beta + \gamma = -\dfrac{(-3)}{1}\\\\\\\implies a – b + a + a + b = 3\\\\\implies 3a = 3\\\end{gathered}

   p(x)=x

   3

   −3x

   2

   +x+1

   A standard cubic equation is of the form ax

   3

   +bx

   2

   +cx+d

   On comparing the two equations, we get

   a=1

   b=−3

   c=1

   d=1

   Given zeroes are:

   α=a−b

   β=a

   γ=a+b

   We know that in a cubic polynomial,

   sum of zeroes = −

   a

   b

   ⟹α+β+γ=−

   1

   (−3)

   ⟹a−b+a+a+b=3

   ⟹3a=3

   \begin{gathered}\implies a = 3 \div 3 = 1\\\\\implies \boxed{\underline{\boxed{\bold{a = 1}}}}\\\\\\\text{We also know that in a cubic polynomial,}\\\\\text{Product of zeroes = $-\dfrac{c}{a}$}\\\\\\\implies \alpha \times \beta \times \gamma = -\dfrac{c}{a}\\\\\\\implies (a-b)(a)(a+b) = -\dfrac{1}{1}\\\\\\\text{Put a = 1}\\\\(1-b)(1)(1+b) = -1\\\\\implies 1^2 – b^2 = -1\\\\\implies 1 – b^2 = -1\\\\\implies b^2 = 1 + 1\\\\\implies b^2 = 2\\\end{gathered}

   ⟹a=3÷3=1

   ⟹

   a=1

   We also know that in a cubic polynomial,

   Product of zeroes = −

   a

   c

   ⟹α×β×γ=−

   a

   c

   ⟹(a−b)(a)(a+b)=−

   1

   1

   Put a = 1

   (1−b)(1)(1+b)=−1

   ⟹1

   2

   −b

   2

   =−1

   ⟹1−b

   2

   =−1

   ⟹b

   2

   =1+1

   ⟹b

   2

   =2

   \implies \boxed{\underline{\boxed{\bold{b = \pm \sqrt{2}}}}}⟹

   b=±

   2

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