can anyone help me solving these theoram questions.........plz.........?

Question

Q3: When ax^2 + bx - 6 is divided by x + 3, the remainder is 9.. Find in terms of a only...the remainder when 2x^3 - bx^2 + 2ax - 4 is divided by x - 2.

Answer 1

when we divide the given expression by X-2
then the remainder will be found by putting 2 in expression
so remainder==2(2)^3-b(2)^2 +2a(2) -4

Answer 2

Let p(x) = ax^2 + bx - 6.

Since it is given that when we divide p(x) by (x + 3), the remainder is 9, we know that p(-3) = 9 (i.e. (x + 3) being a factor implies we plugged in -3 for x in order to get the remainder of 9.

p(-3) = 9, BUT
p(-3) = a(-3)^2 + b(-3) - 6

So we can equate those two.

a(-3)^2 + b(-3) - 6 = 9
a(9) - 3b - 6 = 9
9a - 3b - 6 = 9
9a - 3b = 15, which reduces to

3a - b = 5

In order to find the remainder when 2x^3 - bx^2 + 2ax - 4 is divided by x - 2,
Let q(x) = 2x^3 - bx^2 + 2ax - 4.
To find the remainder, we solve for q(2) [since (x - 2) is a factor, we must solve for q(2) to determine the remainder]

q(2) = 2(2)^3 - b(2)^2 + 2a(2) - 4
q(2) = 16 - 4b + 4a - 4
q(2) = 12 - 4b + 4a

BUT, as solved above, 3a - b = 5, which means b = 3a - 5.
So we substitute each occurrance of b for (3a - 5), to get:

q(2) = 12 - 4b + 4a
q(2) = 12 - 4[3a - 5] + 4a
q(2) = 12 - 12a + 20 + 4a
q(2) = 32 - 8a

And now we've solved this " in terms of a only " as the question said.

EDIT: Made an algebraic error in my first attempt, but fixed it. Thank goodness I did the problem in steps!

Answer 3

Simply divide ax^2 + bx - 6 by x+3
Quotient = ax+(b-3a)
Remainder = -6-3(b-3a)
But the remainder is 9
Simplify and u get :
3a-b = 5
b = 3a-5

Simply divide 2x^3 - bx^2 + 2ax - 4 by x - 2
Quotient=2x^2 + (4-b)x + 2(a-b+4)
Remainder = 4(a-b+4)-4
Put b = 3a-5

Remainder = 32-8a
in terms of a only...the remainder

Answer 4

From the first condition, the remainder is the value of the function at the root of x+3 which is -3:
9 = a(-3)^2-3b-6 or
9a -3b = 15 or upon division by 3
3a -5= b

The second condition says that substituting 2 gives what?
2*2^3-b*2^2+2*a*2-4 = 16
16-4b+4a-4=
12-4b+4a= upon substituting for b from 3a -5 =b
12-4(3a-5)+4a =
12-12a+20+4a=

32-8a

Answer 5

You have a simple way to do the problem.

By remainder theorem,

a(-3)^2 - 3b - 6 = 9
which can be reduced to
3a - b - 5 = 0......(1)

Also, we have
2(2)^3 - 4b + 4a - 4 = r......(2)

Use combination,
-4x(1) + (2): -8a + 32 = r......(3)

Therefore, the remainder is: r = -8a + 32