Question

if (x-2) and (x+3) are factors of x^3+ax^2+bx-30 find a and b​

Answer 1

Solution:

Given,

x-2 is a factor of x³+ax²+bx+16

⇒2 is the zero of polynomial p(x)=x³+ax²+bx+16

⇒p(2)=(2)³+a(2)²+b(2)+16

⇒0=8+4a+2b+16

Given,b=4a

⇒0=8+4a+2(4a)+16

⇒0=8+12a+16

⇒12a= -24

⇒a=(-24)/12

⇒a= -2

b=4a

⇒b=4(-2)

⇒b= -8

value of a is -2 and b is -8

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Answer 2

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Let p(x) = x³ + ax² + bx - 30

Since x-2 and x+3 are the factors of p(x) ,

x-2 = 0 and x+3 = 0

x = 2 and x = -3

2 and -3 are zeroes of p(x)

Hence,

p(2) = 0

p(-3) = 0

p(2) = 2³ + 2²a + 2b - 30 = 0

8 +4a+2b = 30

4+2a+b = 15

2a+b = 11 -------(1)

p(-3) = -3³ -3²a -3b -30 = 0

-27 + 9a - 3b = 30

-9 + 3a - b = 10

3a - b = 19 --------(2)

(1) + (2) =>

2a+b+3a-b = 19+11

5a = 30

a = 6

b = 11 - 2a

b = 11 - 12

b = -1

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