Question

if the zeros of the polynomial x^3-3X^2+ X+1 are a-b,a,a+b,find a and b​

Answer 1

Step-by-step explanation:

your answer is there u may check and verify

Answer 2

Question :-

If the zeros of polynomial x^3 - 3x^2 + x + 1 are a - b, a, a + b, find a and b ?

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Solution :-

Given Information :-

• a polynomial is given, P(x)=$\sf\ x^{3} +3x^{2} + x +1$

• Zeroes = (a-b), a, (a+b)

To Find :-

• We have to find the value of a and b

Calculation :-

$\therefore$The given polynomial is a cubic polynomial and the given zeroes are (a-b),a, (a+b)

Also,

From, P(x) =$\large\sf\ x^3+3x^2+x-1$

• a = 1
• a = 1b = -3
• a = 1b = -3 c = 1

So,

$\large{\boxed{\sf Sum \: of \: zeroes (\alpha \: + \: \beta) = \frac{-b}{a} }}$

$\large{\boxed{\sf product \: of \: Zeroes(\alpha\beta + \beta\gamma + \gamma\alpha)= \frac{c}{a} }}$

Now,

Using this formula,

Using this formula, We get,

$\large::\implies\sf\alpha+ \beta+ \gamma = \frac{-b}{a} \\ \\ \large::\implies\sf\ a-b +a +a+b = -3 \\ \\ \large::\implies\sf\ 3a = -3 \\ \\ \large::\implies\bf\ a= -1$

$\large::\implies \sf\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \\ \\ \large::\implies \sf\ (a-b) a \: + \: a(a+b) \: + \: (a+b)(a-b) = 1 \\ \\ \large::\implies\sf\ a^2-ab + a^2+ab+ a^2-b^2 = 1 \\ \\ \large::\implies\sf\ 3a^2-b^2 =1$

Putting the value of a = -1

$\large::\implies\sf\ 3(-1)^2- b^2 = 1 \\ \\ \large::\implies \sf\ 3 - b^2 = 1 \\ \\ \large::\implies\bf\ -b^2 = -2$

Cancellation of the negative sign on both the sides,

Cancellation of the negative sign on both the sides,We get,

$\large::\implies\bf\ b = ±\sqrt{2}$

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Final Answer :-

• a = 1
• b = ±$\sf\sqrt{2}$

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