Can anyone solve these questions 4 me..plz............That's nt my home work.....?

Question

Q1: Given that 4x^3 - 6x^2 + 1 =(x - 2)(x + 1)Q(x) + ax + b where Q(x) is a polynomial.Find a and b.

Q2: Find the value of the unknown constants for the identities: 3x^2 + 6x - 4= p(x - q)^2 + r.
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.

Q4: The expressions 8x^3 + ax^2 + bx - 9 leaves remainders -95 and 3 when divided by x + 2 and 2x - 3 respectively. Calculate the value of a and b.


PLZ SOLVE THEM...i WILL BE VERY THANKFUL 2 U...........................tHE FOLLOWING QUESTIONS ARE REGARDING REMAINDER AND FACTOR THEORAMS

Answer 1

Q1: Given that 4x^3 - 6x^2 + 1 =(x - 2)(x + 1)Q(x) + ax + b where Q(x) is a polynomial.Find a and b.
(x-2)(x+1)= x^2 -x -2
Using long division divide 4x^3 -6x^2 +0x +1 by x^2-x-2
You will get 4x-2 with a remainder of 6x-3
Hence a = 6 and b = -3

Q2: Find the value of the unknown constants for the identities: 3x^2 + 6x - 4= p(x - q)^2 + r.
x^2 +2x -4/3 = p/3(x-q)^2 +r/3
x^2 +2x +1 -4/3 =p/3(x-q)^2 +1
(x+1)^2= p/3(x-q)^2 + 7/3
so if p=3, and q= -1, and r= 7, we get
(x-1)^2 +7/3

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.

Divide ax^2+bx-6 by x+3 and get a remainder of -3ax-3b-6
Set this = 9 getting ax+b= -5.or b= -5-ax.
The remainder when you divide 2x^3-bx^2 +2ax -4 by x-2 is
2a-8b +12 = 2a-8(-5-ax)+12 = 2a+40 +8ax +12
=8ax+2a+52 = 2a(4x+1)+52


Q4: The expressions 8x^3 + ax^2 + bx - 9 leaves remainders -95 and 3 when divided by x + 2 and 2x - 3 respectively. Calculate the value of a and b.

Dividing by x+2 gives remainder of -2b+4a-73=-95, so
-2b+4a = -22 or 2a-b =-11*
For the other remainder I get 1.5b+ 2.25a+54= 3, so
2.25a +1.5b =-51**
Solve * and ** simultaneously to get a and b

Answer 2

Q1: Given that 4x^3 - 6x^2 + 1 =(x - 2)(x + 1)Q(x) + ax + b where Q(x) is a polynomial.Find a and b.
(x-2)(x+1)= x^2 -x -2
Using long division divide 4x^3 -6x^2 +0x +1 by x^2-x-2
You will get 4x-2 with a remainder of 6x-3
Hence a = 6 and b = -3

Q2: Find the value of the unknown constants for the identities: 3x^2 + 6x - 4= p(x - q)^2 + r.
x^2 +2x -4/3 = p/3(x-q)^2 +r/3
x^2 +2x +1 -4/3 =p/3(x-q)^2 +1
(x+1)^2= p/3(x-q)^2 + 7/3
so if p=3, and q= -1, and r= 7, we get
(x-1)^2 +7/3

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.

Divide ax^2+bx-6 by x+3 and get a remainder of -3ax-3b-6
Set this = 9 getting ax+b= -5.or b= -5-ax.
The remainder when you divide 2x^3-bx^2 +2ax -4 by x-2 is
2a-8b +12 = 2a-8(-5-ax)+12 = 2a+40 +8ax +12
=8ax+2a+52 = 2a(4x+1)+52


Q4: The expressions 8x^3 + ax^2 + bx - 9 leaves remainders -95 and 3 when divided by x + 2 and 2x - 3 respectively. Calculate the value of a and b.

Dividing by x+2 gives remainder of -2b+4a-73=-95, so
-2b+4a = -22 or 2a-b =-11*
For the other remainder I get 1.5b+ 2.25a+54= 3, so
2.25a +1.5b =-51**
Solve * and ** simultaneously to get a and b

Answer 3

x² = x+6 x² - x - 6 = 0 (x - 3)(x + 2) = 0 So x = 3 or - 2 x² + 2x = 0 x(x + 2) = 0 So x = 0 or - 2 60 = x² - 4x x²- 4x - 60 = 0 (x - 10)(x + 6) = 0 So x = 10 or -6 4x = x² - 40 5 x² - 4x - 40 5 = 0 (x - 9)(x + 5) = 0 So x = 9 or -5 -6x = x² + 9 x² + 6x + 9 = 0 (x + 3)(x + 3) = 0 (x + 3)² = 0 So x = -3 -15=x² + 8x x² + 8x +15 = 0 (x + 5)(x + 3) = 0 So x = -5 or -3 -9x = x² + 14 x² + 9x + 14 = 0 (x + 7)(x + 2) = 0 So x = -7 or - 2 desire that facilitates. =D

Answer 4

Q1: The rt hand side (RHS) of your equation is the divisor [ (x-2)(x+1)] times the quotient [ Q(x) ] added to a remainder of 2 terms. The left hand side (LHS) is the dividend in your division operation. So expand your divisor: x^2 - x - 2 and divide into your dividend (LHS) and get your quotient [ Q(x) = 4x - 2 ] and your remainder:

6x - 3 ( making a = 6 and b = -3.)

That should get you started.

Q2:
Expand RHS: [px^2 - 2pqx + pq^2 ] + r

In order for right polynomial to = left, coefficients of same power terms must be equal. That means

p must = 3 and -2pq = 6 so
q must = -1.
Finally (the coefficient of x^0) on the LHS is -4 and on the right is [pq^2 + r] or [3(-1)^2 + r] or [3 + r] so r must = -7

Answer 5

ummm 2

Answer 6

Wow, that's not your homework?... Sure looks like it to me...

If it's not yours, it's someone else's who conned you into doing his or her homework for them