Q. If one of the zeroes of the cubic polynomial x3 + ax² + bx + c is -1, then the product of the other two zeroes is

(a) Let p(x) = x3 + ax2 + bx + c

Let a, p and y be the zeroes of the given cubic polynomial p(x).

∴  α = -1                                        [given]

and p(−1) = 0

⇒ (-1)3 + a(-1)2 + b(-1) + c = 0

⇒ -1 + a- b + c = 0

⇒ c = 1 -a + b                                                            …(i)

We know that,

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refer to attatchment#1

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αβγ = -c

⇒ (-1)βγ = −c                                                                             [∴α = -1]

⇒ βγ = c

⇒ βγ = 1 -a + b                                                                [from Eq. (i)]

Hence, product of the other two roots is 1 -a + b.

Alternate Method

Since, -1 is one of the zeroes of the cubic polynomial f(x) = x2 + ax2 + bx + c i.e., (x + 1) is a factor of f{x).

Now, using division algorithm,

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refer to attatchment #2

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⇒x3 + ax2 + bx +c = (x + 1) x {x2 + (a – 1)x + (b – a + 1)> + (c – b + a -1)

⇒x3 + ax2 + bx + (b – a + 1) = (x + 1) {x2 + (a – 1)x + (b -a+ 1)}  

Let a and p be the other two zeroes of the given polynomial, then

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refer to attatchment #3

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the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: