WRITE THE POLYNOMIAL AS THE PRODUCT OF TWO BINOMIALS

Write each expression as the product of two binomials.

Problem 1 :

y(y + 1) – 1(y + 1)

Solution :

The polynomial is having binomial factor of y + 1.

Factor out (y + 1)

= (y - 1) (y + 1)

Problem 2 :

3b(b - 2) – 4(b - 2)

Solution :

The polynomial is having binomial factor of b - 2.

Factor out (b – 2)

= (3b - 4) (b - 2)

Problem 3 :

2x(y + 4) + 3(y + 4)

Solution :

The polynomial is having binomial factor y + 4.

Factor out (y + 4)

= (2x + 3) (y + 4)

Problem 4:

a³ - 3a² + 3a – 9

Solution :

= a³ - 3a² + 3a – 9

= a² (a - 3) + 3(a - 3)

= (a² + 3) (a - 3)

Problem 5 :

2x³ - 3x² - 4x + 6

Solution :

= 2x³ - 4x - 3x² + 6

= 2x (x² - 2) – 3 (x² - 2)

= (2x – 3) (x² - 2)

Problem 6 :

y³ + y² - 5y – 5

Solution :

= y³ - 5y + y² – 5

= y(y² - 5) + 1(y² - 5)

= (y + 1) (y² - 5)

Problem 7:    

x² + 7x + x + 7

Solution :

= x² + 7x + x + 7

= x(x + 7) + 1(x + 7)

= (x + 1) (x + 7)

Problem 8 :

x² + 5x + 6

Solution :

 x² + 5x + 6

Split the middle term into two term

= x² + 2x + 3x + 6

By grouping,

= (x² + 2x) + (3x + 6)

By taking the common factor, we get

= x(x + 2) + 3(x + 2)

= (x + 2) (x + 3)

Problem 9 :

x² - x – 6

Solution :

x² - x – 6

Split the middle term into two term

= x² - 3x + 2x – 6

By grouping,

= (x² - 3x) + (2x – 6)

By taking the common factor, we get

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

= (x + 2) (x - 3)

Problem 10 :

x² + 9x + 20

Solution :

x² + 9x + 20

Split the middle term into two term

= x² + 4x + 5x + 20

By grouping,

= (x² + 4x) + (5x + 20)

By taking the common factor, we get

= x(x + 4) + 5(x + 4)

= (x + 4) (x + 5)

Problem 11 :

 x² − 9

Solution :

= x² − 9

= x² − 3²

Using algebraic identity,

a² − b² = (a - b)(a + b)

x² − 3² = (x - 3)(x + 3)

Problem 12 :

y² − 100

Solution :

= y² − 100

= y² − 10²

Using algebraic identity,

a² − b² = (a - b)(a + b)

y² − 10² = (y - 10)(y + 10)

Problem 13 :

z² + 6z + 9

Solution :

= z² + 6z + 9

= z² + 3z + 3z + 9

Factroing z from first two terms and 3 from the third and fourth terms, we get

= z(z + 3) + 3(z + 3)

= (z + 3)(z + 3)

Problem 14 :

m² + 16m + 64

Solution :

= m² + 16m + 64

= m² + 8m + 8m + 64

= m(m + 8) + 8(m + 8)

= (m + 8)(m + 8)

Problem 15 :

x² − 3x + 4ax − 12a

Solution :

= x² − 3x + 4ax − 12a

Factoring x from the first two terms and 4a from the third fourth terms

= x(x - 3) + 4a(x - 3)

= (x - 3)(x + 4a)

Problem 16 :

n³ − 9n

Solution :

= n³ − 9n

Factoring n, we get

= n(n² − 9)

= n(n² − 3²)

= n(n - 3)(n + 3)

Problem 17 :

You are installing new tile on an outside patio. The area (in square feet) of the rectangular patio can be represented by 8x² + 33x + 4. Write the expressions that represent the dimensions of the patio

Solution :

= 8x² + 33x + 4

= 8x² + 32x + 1x + 4

Factoring 8x from the first two terns and factoring 1 from third and fourth terms, we get

= 8x(x + 4) + 1(x + 4)

= (8x + 1)(x + 4)

length = 8x + 1 and width = x + 4.

Problem 18 :

The length of a rectangular game reserve is 1 mile longer than twice the width. The area of the reserve is 55 square miles. What is the width of the reserve?

Solution :

Use the formula for the area of a rectangle to write an equation for the area of the reserve. Let w represent the width. Then 2w + 1 represents the length. Solve for w.

w(2w + 1) = 55

2w² + w = 55

2w² + w − 55 = 0

2w² - 10w + 11w − 55 = 0

2w(w - 5) + 11(w - 5) = 0

(2w + 11)(w - 5) = 0

2w + 11 = 0 and w - 5 = 0

w = -11/2 and w = 5

A negative width does not make sense, so you should use the positive solution. So, the width of the reserve is 5 miles