Question

x³+ax²+bx+6 has (x-2) as factor and leaves remainder 3 when divided by (x-3), find values of a and b​

Answer 1

Step-by-step explanation:

this is your answer....

Answer 2

P(x) = x^3 + ax^2 + bx + 6

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Reminder :

x - a = 0 ==》 x = a

P(a) = Reamaining of the diving P(x) by

( x - a )

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x - 2 = 0 ==》x = 2 ==》P(2) = 0

(2)^3 + a(2)^2 + b(2) + 6 = 0

8 + 4a + 2b + 6 = 0

4a + 2b + 14 = 0

4a + 2b + 14 - 14 = 0 - 14

4a + 2b = - 14

( 4a + 2b ) ÷ 2 = ( -14 ) ÷ 2

2a + b = - 7 ( ♡ )

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x - 3 = 0 ==》 x = 3 ==》P(3) = 3

(3)^3 + a(3)^2 + b(3) + 6 = 3

27 + 9a + 3b + 6 = 3

9a + 3b + 33 = 3

9a + 3b + 33 - 33 = 3 - 33

9a + 3b = - 30

( 9a + 3b ) ÷ 3 = ( - 30 ) ÷ 3

3a + b = - 10 ( ☆ )

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( ♡ ) , ( ☆ ) in pair system of equation

( ☆ ) : 3a + b = - 10 3a + b = - 10

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( ♡ ) : 2a + b = - 7 × (-1) - 2a - b = + 7

___________ +

a = - 3

So :

2( - 3 ) + b = - 7

- 6 + b = - 7

- 6 + 6 + b = - 7 + 6

b = - 1

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Thus : a = - 3 & b = - 1