Question

if alpha beta gamma are the zeros of the polynomial Y is square minus 6 y square minus y + 30 then the value of alpha beta gamma​

Answer 1

The values of $\alpha,\beta,\gamma$ are $-2,3,5$

Step-by-step explanation:

Given

$\alpha,\beta,\gamma$ are zeroes of polynomial

$y^3-6y^2-y+30$

To find out:

Values of $\alpha,\beta,\gamma$

Solution:

Let

$p(y)=y^3-6y^2-y+30$

Putting y = 1

$p(1)=1^3-6\times 1^2-1+30=1-6-1+30=24\ne 0$

Putting y = -1

$p(-1)=(-1)^3-6\times (-1)^2-(-1)+30=-1-6+1+30=24\ne 0$

Putting y = -2

$p(-2)=(-2)^3-6\times (-2)^2-(-2)+30=-8-24+2+30=32-32=0$

Since on putting y = -2, the value of polynomial p(y) is zero

Therefore, y = -2 is a zero of the polynomial p(y)

Therefore, y+2 is a factor of polynomial p(y)

Thus, p(y) can be written as

$p(y)=(y+2)(y^2-8y+15)$

$\implies p(y)=(y+2)(y^2-5y-3y+15)$

$\implies p(y)=(y+2)[y(y-5)-3(y-5)]$

$\implies p(y)=(y+2)(y-3)(y-5)$

Thus,

(y+2), (y-3), (y-5) are factors of polynomial p(y)

Or, y = -2, y = 3, y = 5 are zeroes of polynomial p(y)

Therefore, the values of $\alpha,\beta,\gamma$ are $-2,3,5$

Hope this answer is helpful.

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