Question

. If - 1 and 2 are the zeroes of the cubic polynomial,
x3 – 2 x2 + ax + b , then find the values of a and b.

Answer 1

Answer: a = -1, b = 2

Step-by-step explanation:

Put  -1 and 2 in the given equation $x^{3} - 2x^{2} + ax+b = 0$

Put x = -1 => -1-2-a+b = 0 => -3-a+b = 0 => b-a = 3 ........(1)

Put x = 2 => 8-8+2a+b = 0 => 2a+b = 0...............(2)

From (1) and (2), a  = -1, b = 2

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Answer 2

Answer:

Thus a=1 and b=-2

Step-by-step explanation:

Let P(x)=x³ – 2 x² + ax + b

-1 is a zero

so p(-1)=0

(-1)³-2(-1)²-a+b=0

a-+b=-1, b=-a-1

now 2 is zero of p(x)

So 2³-2*2²+2a+b=0

8-8+2a-1-1=0

2a=2, a=1

b=-1-1=02

Thus a=1 and b=-2