FIND THE COEFFICIENTS FROM THE PRODUCT OF ALGEBRAIC TERMS

Problem 1 :

2.3x × 1.1x × 2.5x, what is coefficient of the product

a)  4.325      b)  6.325      c)  3.325      d)  None of these

Solution :

2.3x × 1.1x × 2.5x

Product of Coefficient = 2.3 × 1.1 × 2.5

= 6.325

So, option (b) is correct.

Problem 2 :

3x + 5yz + 7x²

write the product of coefficients in this expression.

Solution :

Expression 7x² + 5yz + 3x

Coefficients 7, 5, 3

Product of coefficients = 7 × 5 × 3

= 105

Problem 3 :

Find the product of 

(4.1X + 5.6Y) (1.2XY - 3Y)

and write all the coefficients.

Solution :

= (4.1X + 5.6Y) (1.2XY - 3Y)

= 4.1X (1.2XY - 3Y) + 5.6Y (1.2XY - 3Y)

= 4.92X²Y - 12.3XY + 6.72XY² - 16.8Y²

= 4.92X²Y + 6.72XY² - 12.3XY - 16.8Y²

Coefficients = 4.92, 6.72, -12.3, -16.8

Problem 4 :

6x = x - 3x(2n - 1)

In the given equations, n is constant. The equation has infinitely many solutions. What is the value of n.

a)  -2/3     b)  -1/3     c)  4/3    d)  5/3

Solution :

6x = x - 3x(2n - 1)

6x = x - 3x(2n) - 3x(-1)

6x = x - 6nx + 3x

6x = 4x - 6nx

6x - 4x = -6nx

2x = -6nx

2 = -6n

n = -2/6

n = -1/3

So, option b is correct.

Problem 5 :

2(x + k) = 2x + 2

In the given equation, k is a constant greater than 1. How many solutions does the eqauation have ?

a) zero   b) Exactly one   c)  Exactly two    d) Infinitely many

Solution :

2(x + k) = 2x + 2

2x + 2k = 2x + 2

2k = 2

The equation will be satisfied when k = 1

When value of k is greater than 14, for example 2

2x - 2x + 4 = 2

4 not equal to 2

Then it may have zero solution. Option a is correct.

Problem 6 :

12x + 16nx = 20

In the given equation, n is constant. The equation has no solution. What is the value of n ?

Solution :

12x + 16nx = 20

For an equation to have no solution, both sides must be equivalent except for the constants. Since there is no x term on the right sides of the equation we need to find the value of n such that the x terms on the left side of the equation cancel out. In other words, the coefficients of x on the left should sum of zero.

12 + 16n = 0

16n = -12

n = -12/16

n = -3/4

Problem 7 :

(x - 4)² - a = x² + bx + b

In the given equation, a and b are constants. Of the equation is true for all values of x, what is the value of a ?

Solution :

(x - 4)² - a = x² + bx + b

x² - 2x(4) + 4² - a = x² + bx + b

x² - 8x + 16 - a = x² + bx + b

By equating the coefficients of corresponding terms, we get

-8 = b

Equating the constants,

16 - a = b

Applying the value of b, we get

16 - a = -8

a = 16 + 8

a = 24

Problem 8 :

The expressions bx² + 11x  and x(4x + a) - 3x where a and b are constants are equivalent, what is the value of a + b ?

a)  12     b)  15    c)  18    d)  21

Solution :

bx² + 11x  = x(4x + a) - 3x

bx² + 11x  = 4x² + ax - 3x

bx² + 11x  = 4x² + (a - 3)x

Equating the coefficients of x²

b = 4

Equating the coefficients of x, we get

11 = a - 3

a = 11 + 3

a = 14

a + b = 14 + 4

= 18

So, the value of a and b is 18.

Problem 9 :

The expression (ax + 5)(bx - 5), where a and b are constants, can be rewritten as cx² + 15x - 25 where c is constant. What is the value of b - a ?

Solution :

(ax + 5)(bx - 5) = cx² + 15x - 25

ax(bx) +ax(-5) + 5(bx) + 5(-5) = cx² + 15x - 25

abx² +(-5a + 5b) x - 25 = cx² + 15x - 25

Equating the coefficients of x2 and x, we get

ab = c

-5a + 5b = 15

-a + b = 3

b - a = 3

So, the value of b - a is 3

Problem 10 :

The expression 12x² - mx - 20 where m is constant, is equivalent to (kx - 5)(nx + 4) where k and n are integer constants. Which of the following must be an integer.

a)  12/m        b)  12/n        c)  m/k         d)  m/n

Solution :

(kx - 5)(nx + 4) = 12x² - mx - 20

kx(nx) + kx(4) - 5(nx) - 5(4) = 12x² - mx - 20

knx² + 4kx - 5nx - 20 = 12x² - mx - 20

knx² + (4k - 5n)x - 20 = 12x² - mx - 20

Equating the coefficients of x² and x, we get

kn = 12 and 4k - 5n = -m

k = 12/n

So, option b is correct.