# Cn anyone solve these...?

the polynomial x^3 + ax^2 + bx - 3 leaves remainders of 27 when divided by x - 2 and a remainder of 3 when divided by x + 1. Calculate the remainder when the polynomial is divided by x - 1.

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• Let p(x) = x^3 + ax^2 + bx - 3

Since we get a remainder when dividing p(x) by (x - 2), it follows that p(2) = 27. However, p(2) is also equal to what is below:

p(2) = 2^3 + a(2)^2 + b(2) - 3

p(2) = 8 + 4a + 2b - 3

p(2) = 5 + 4a + 2b

So now we can equate this to 27, getting

27 = 5 + 4a + 2b

22 = 4a + 2b, and reducing this, we get

11 = 2a + b

We also know we get a remainder of 3 when dividing p(x) by (x + 1), so we know that p(-1) = 3. But, p(-1) is also equal to:

p(-1) = (-1)^3 + a(-1)^2 + b(-1) - 3

p(-1) = -1 + a - b - 3

p(-1) = -4 + a - b

Equating this to 3, we get

3 = -4 + a - b

7 = a - b

Two equations, two unknowns:

11 = 2a + b

7 = a - b

To solve this, we can use elimination and add the two equations.

18 = 3a

Therefore, a = 6

Plugging this into the second equation,

7 = 6 - b, therefore,

1 = -b, so b = -1

So a = 6 and b = -1

• By remainder theorem,

2^3+a2^2+2b - 3 = 27

which can be reduced to

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

(-1)^3+a-b - 3 = 3

which can be reduced to

a-b = 7......(2)

Solving the system of (1) and (2) gives,

a = 6, b = -1

Therefore, the remainder when the polynomial is divided by x-1 is: 1+6-1-3 = 3

• using the remainder theorem

f(2)=2^3+a(2)^2+b(2)-3=27

so 4a+2b=22 (1)

f(-1)=(-1)^3+a(-1)^2+b(-1)-3=3

so a-b=7 (2)

(2)*2

2a-2b=14

4a+2b=22

6a=36

dividing by 6

a=6

aub in (2)

6-b=7

-b=1

b=-1

so the poly is

x^3+6x^2-b-3

f(1) =1+6-1-3

=3

so the remainderwhen divided by x-1 is 3