## 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

Question

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.

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2021-07-22T14:25:18+00:00
2021-07-22T14:25:18+00:00 1 Answers
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## Answers ( )

Answer:(x – 1)(3x + 2)

Step-by-step 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