Question

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

Answer 1

$\huge{\bf{\underline{\underline{AnSwEr}}}}$

GivEn:-

• 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

StEp-By-step-explanation:-

$\therefore$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

• b = -3

• c = 1

So,

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

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

Now, using this formula;

$\large\leadsto\tt\alpha+ \beta+ \gamma = \frac{-b}{a} \\ \\ \large\leadsto\tt\ a-b +a +a+b = -3 \\ \\ \large\leadsto\tt\ 3a = -3 \\ \\ \large\leadsto\tt\ a= -1$

$\rule{250}{2}$

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

Putting the value of a = -1

$\large\leadsto\tt\ 3(-1)^2- b^2 = 1 \\ \\ \large\leadsto\tt\ 3 - b^2 = 1 \\ \\ \large\leadsto\tt\ -b^2 = -2$

Cancel,the negative sign both side

$\large\leadsto\tt\ b= \sqrt{2}$