Question

Given that the zeroes of the cubic polynomial

x

3

– 15x

2

+ 66x – 80 are of the form a, a + b, a +

2b for some positive integers a and b, find the

values of a and b.​

Answer 1

Answer:

remaining part is given below

a= 2, b= 3

and zeroes are (2,5,8)

Please mark my answer as brainliest

Answer 2

Answer:

Step-by-step explanation:

Given:

${x}^{3} - 15{x}^{2} + 66x -80$

zeroes of cubic polynomial is in the form a,a+b and a+2b .

To find: Find the value of a and b.

Solution:We know that zeroes of cubic polynomial and coefficients of cubic polynomial have some relation.

If $\alpha ,\beta ,\gamma$are the zeroes of cubic polynomial

$a{x}^{3} + b {x}^{2} + cx + d$

then

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} \\ \\ \alpha \beta \gamma = \frac{ - d}{a}$

So, here zeros of polynomial are a,a+b and a+2bapply the relation

$a + a + b + a + 2b = 15 \\ \\ or \\ \\ 3a + 3b = 15 \\ \\ or \\ \\\bold{ a + b = 5}...eq1$

From last relation

$a(a + b)(a + 2b) = 80 \\ \\ a(5)(a+b+b ) = 80 \\ \\5a(5+b)=80\\\\\text{put value from eq1}\\\\ 5a (5+5 - a) = 80 \\ \\ 5a (10 - a) = 80 \\ \\or\\\\a(10-a)=16\\ \\ 10a - {a}^{2} -16= 0 \\ \\ or \\ \\ {a}^{2} -10a +16 = 0 \\ \\ {a}^{2} - 8a - 2a +16 = 0 \\ \\ a(a - 8) -2(a - 8) = 0 \\ \\ (a - 8)(a -2) = 0 \\ \\ either \\ \\ a = 8\\ \\ or \\ \\ a = 2$

Put a=8 in eq1$8 + b = 5 \\ \\ b = - 3$

put a=2 in eq1$2 + b = 5 \\ \\ b = 3$

ATQ values of a and b are some positive real values. So discard negative value of b and that corresponding value of a.

Final answer:

$\boxed{\bold{a = 2 ,\: b = 3}}$

Hope it helps you.

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