HOW TO FIND THE MISSING FACTOR OF A POLYNOMIAL

Find the missing factor :

Problem 1 :

4 × ? = 8x

Solution :

4 × ?

= 4 × 2x

= 8x

So, the missing factor is 2x.

Problem 2 :

5 × ? = 15y

Solution :

5 × ?

= 5 × 3y

= 15y

So, the missing factor is 3y.

Problem 3 :

3 × ? = 9a²

Solution :

3 × ?

= 3 × 3a²

= 9a²

So, the missing factor is 3a².

Problem 4 :

3x² × ? = 12x²

Solution :

3x² × ?

= 3x² × 4

= 12x²

So, the missing factor is 4.

Problem 5 :

? × 7y = 7y²

Solution :

? × 7y

= y × 7y

= 7y²

So, the missing factor is y.

Problem 6 :

? × 2a = -8a

Solution :

? × 2a

= -4 × 2a

= -8a

So, the missing factor is -4.

Problem 7 :

p × ? = -pq

Solution :

p × ?

= p × -q

= -pq

So, the missing factor is -q.

Problem 8 :

? × 3a = 6a³

Solution :

? × 3a

= 2a² × 3a

= 6a³

So, the missing factor is 2a².

Problem 9 :

8s × ? = -24st

Solution :

8s × ?

= 8s × -3t

= -24st

So, the missing factor is -3t.

Problem 10 :

Factorize x² - 6x - 16

Solution :

x² - 6x - 16

= x² - 8x + 2x - 16

= x(x - 8) + 2(x - 8)

= (x - 8)(x + 2)

Problem 11 :

Factorize x² - 10xy + 24y²

Solution :

= x² - 10xy + 24y²

= x² - 6xy - 4xy + 24y²

= x(x - 6y) - 4y(x - 6y)

= (x - 4y)(x - 6y)

Problem 12 :

Factorize 125x³ - 1

Solution :

= 125x³ - 1

=  5³x³ - 1

= (5x)³ - 1³

a³ - b³ = (a - b) (a² + ab + b²)

= (5x - 1)( (5x)² + (5x)(1) + 1²)

= (5x - 1)(25x² + 5x + 1)

Problem 13 :

Factorize a³  - 14a² + 49a

Solution :

= a³  - 14a² + 49a

= a(a²  - 14a + 49)

= a(a²  - 7a - 7a + 49)

= a[a(a - 7) - 7(a - 7)]

= a(a - 7) (a - 7)

Problem 14 :

The Parthenon in Athens, Greece, is an ancient structure that has a rectangular base. The length of the base of the Parthenon is 8 meters more than twice its width. The area of the base is about 2170 square meters. Find the length and width of the base.  

Solution :

Let x be the width of the rectangular base.

Length = 2x + 8

Area of the base = 2170

x(2x + 8) = 2170

2x² + 8x = 2170

2x² + 8x - 2170 = 0

x² + 4x - 1085 = 0

x² + 35x - 31x - 1085 = 0

x(x + 35) - 31(x + 35) = 0

(x - 31)(x + 35) = 0

x = 31 and x = -35

So, width of the rectangle is 31 m 

length = 2(31) + 8

= 62 + 8

= 70 m

Problem 14 :

Your friend says that to solve the equation 5x² + x − 4 = 2, you should start by factoring the left side as (5x − 4)(x + 1). Is your friend correct? Explain.

Solution :

5x² + x − 4 = 2

5x² + x − 4 - 2 = 0

5x² + x − 6 = 0

5x² + 6x - 5x − 6 = 0

x(5x + 6) - 1(5x + 6) = 0

(x - 1)(5x + 6)

The factors are (x - 1)(5x + 6), but he given factors are (5x − 4)(x + 1). So, it is incorrect.

Problem 15 :

The length of a rectangular birthday party invitation is 1 inch less than twice its width. The area of the invitation is 15 square inches. Will the invitation fit in the envelope shown without being folded? Explain.

Solution :

Let x be the width.

Length = 2x - 1

Area of the invitation = 15 square inches

x(2x - 1) = 15

2x² - x = 15

2x² - x - 15 = 0

2x² - x - 15 = 0

2x² - 6x + 5x - 15 = 0

2x(x - 3) + 5(x - 3) = 0

(2x + 5)(x - 3) = 0

x = -5/2 and x = 3

Width = 3 inches

length = 2(3) - 1

= 6 - 1

= 5 inches

Length and width of envelop is smaller than than folder, then it can be fixed.

Problem 16 :

The length of a rectangle is 1 inch more than twice its width. The value of the area of the rectangle (in square inches) is 5 more than the value of the perimeter of the rectangle (in inches). Find the width.

Solution :

Let x be the width

Length = 2x + 1

Area of rectangle = x(2x + 1)

Perimeter of the rectangle = 2(length + width)

= 2(x + 2x + 1)

= 2(3x + 1)

x(2x + 1) + 5 = 2(3x + 1)

2x² + x + 5 = 6x + 2

2x² + x - 6x + 5 - 2 = 0

2x²  - 5x + 3 = 0

2x²  - 2x - 3x + 3 = 0

2x(x - 1) - 3(x - 1) = 0

(2x - 3)(x - 1) = 0

x = 3/2 and x = 1

The possible widths are 1 or 1.5