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A

The expression ax²+bx+c is divisible by x-1, has a
remainder of 2 when divided by x+1 and has a remainder of 8 and when divided by x-2 find the values of a, b, and c and
factorise completely.​

User Amit Ray
by
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1 Answer

1 vote

Answer:

(x - 1)(3x + 2)

Explanation:

By the Remainder theorem

Given expression is divisible by x - 1 then f(1) = 0

when divided by (x + 1) then f(- 1) = 2

when divided by x - 2 then f(2) = 8

Substitute x = 1, x = - 1, x = 2 into the expression , that is

a + b + c = 0 → (1)

a - b + c = 2 → (2)

4a + 2b + c = 8 → (3)

Subtract (1) from (2) to eliminate a and c

- 2b = 2 ( divide both sides by - 2 )

b = - 1

Subtract (1) from (3) to eliminate c

3a + b = 8 ← substitute b = - 1

3a - 1 = 8 ( add 1 to both sides )

3a = 9 ( divide both sides by 3 )

a = 3

Substitute a = 3, b = - 1 into (1) and evaluate for c

3 - 1 + c = 0

2 + c = 0 ( subtract 2 from both sides )

c = - 2

Then a = 3, b = - 1, c = - 2

Thus

ax² + bx + c

= 3x² - x - 2

= (x - 1)(3x + 2) ← in factored form