WRITING POLYNOMIALS IN DIVISOR TIMES QUOTIENT PLUS REMAINDER FORM

Problem 1 :

The equation 

kx2+14x-203x-2 = 5x+8 - 43x-2

is true for all values of x ≠ 2/3, where k is a constant. What is the value of k ?

(a)  8       (b)  9     (c)  11     (d)  15

Solution :

kx2+14x-203x-2 = 5x+8 - 43x-2=(5x+8)(3x-2)-43x-2=15x2-10x+24x-16-43x-2kx2+14x-203x-2 =15x2+14x-203x-2

So, the value of k is 15.

Problem 2 :

3x2+x+2x-1 = ax+b + cx -1

Where a, b and c are constants. What is the value of a + b + c ?

Solution :

2x2-5x+2x-1 = ax+b + cx -1 =(ax+b)(x-1)+cx -1=ax2-ax+bx-b+cx -13x2+x+2x-1=ax2+x(b-a)+(c-b)x -1

Comparing the coefficient of x2, x and constants.

a = 3

b - a = 1  ----(1)

c - b = 2 -----(2)

Applying the value of a in (1), we get

b - 3 = 1

b = 4

Applying the value of b in (2), we get

c - 4 = 2

c = 6

a + b + c = 3+4+6

= 13

So, the answer is 13.

Problem 3 :

In the following find the value of R

2x2-5x+32x+1 = x-3 + R2x +1

Solution :

2x2-5x+32x+1 = x-3 + R2x +1= (x-3)(2x+1)+R2x +1= 2x2+x-6x-3+R2x +1= 2x2-5x-3+R2x +1= 2x2-5x+(R-3)2x +1

Comparing constants :

3 = R - 3

R = 3 + 3

R = 6

Alternative way :

dividingpolywiqr
2x2-5x+32x+1 = x-3 + 62x+1

So, the value of R is 6.

Problem 4 :

14x2+9x-20ax-1 = 7x+8 + -12ax-1

In the equation above, a is a constant and ax − 1 ≠ 1. What is the value of a ?

Solution :

14x2+9x-20ax-1 = 7x+8 + -12ax-1= (7x+8)(ax - 1) -12ax-1= 7ax2-7x+8ax-8 -12ax-1= 7ax2+x(-7+8a)-20ax-1

Comparing the coefficients of x, we get

-7 + 8a = 9

8a = 9 + 7

8a = 16

a = 2

Problem 5 :

From the information given below, find the value A.

6x2-5x+4-3x+1 = -2x+1+A-3x+1

Solution :

6x2-5x+4-3x+1 = -2x+1+A-3x+1= (-2x+1)(-3x+1)+A-3x+1=6x2-2x-3x+1+A-3x+1=6x2-5x+ (1+A)-3x+1

Comparing constants, we get

4 = 1 + A

A = 3

Problem 6 :

In the following, find the value of a.

x2-x-ax-2 = x+1+8x-2

Solution :

x2-x-ax-2 = x+1+8x-2= (x+1)(x-2)+8x-2= x2-2x+1x-2+8x-2= x2-x-2+8x-2= x2-x+6x-2

Comparing constants, the value of a is 6.

Problem 7 :

In the following, find the value of A.

-2x2+5x+10-x+4 = 2x+3-A-x+4

Solution :

-2x2+5x+10-x+4 = 2x+3-A-x+4= (2x+3)(-x+4)-A-x+4= -2x2+8x-3x+12-A-x+4= -2x2+5x+(12-A)-x+4

Comparing the constant terms, we get

10 = 12 - A

A = 12 - 10

A = 2

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