SIMPLIFYING POLYNOMIALS WITH DIVISION PRACTICE WORKSHEET

Simplify the following :

(1)  [(x²-2x)/(x+2)] ⋅ [(3x+6)/(x-2)]

(2)  [(x²-81)/(x²-4)] ⋅ [(x²+6x+8)/(x²-5x-36)]

(3)  [(x²-3x-10)/(x²-x-20)] ⋅ [(x²-2x+4)/(x³+8)]

(4)  [(x²-16)/(x^2-3x+2)] ⋅ [(x²-4)/(x³+64)] ⋅ 

[(x²-4x+16)/(x²-2x-8)]

(5)  [(3x²+2x-1)/(x²-x-2)] [(2x²-3x-2)/(3x²+5x-2)]

(6)  [(2x-1)/(x²+2x+4)] ⋅[(x⁴-8x)/(2x²+5x-3)] ⋅

[(x+3)/(x²-2x)]

(7)  [(a+b)/(a-b)] [(a³-b³)/(a³+b³)]

(8)  [(x²-9y²)/(3x-3y)] ⋅ [(x²-y²)/(x²+4xy+3y²)]

(9)  [(x²-4x-12)/(x²-3x-18)] ⋅ [(x²-2x-3)/(x²+3x+2)]

(10)  [(x²-3x-10)/(x²-x-20)]⋅[(x²-4x+16)/(x³+64)]

(11)   [(x²-16)/(x-2)] [(x²-4)/(x³+64)]

(12)  [(x + 7)/(x²+14x+49)] [(x²+8x+7)/(x+1)]

Detailed Answer Key

Problem 1 :

[(x²-2x)/(x+2)] ⋅ [(3x+6)/(x-2)]

Solution :

Let f(x)  =   [(x²-2x)/(x+2)] ⋅ [(3x+6)/(x-2)]

f(x)  =  [(x²-2x)/(x+2)] ⋅ [(3x+6)/(x-2)]

f(x)  =  [x(x-2)/(x+2)] ⋅ [3(x+2)/(x-2)]

f(x)  =  3x

So, the value of f(x) is 3x.

Problem 2 :

[(x²-81)/(x²-4)] ⋅ [(x²+6x+8)/(x²-5x-36)]

Solution :

Let f(x)  =  [(x²-81)/(x²-4)] ⋅ [(x²+6x+8)/(x²-5x-36)]

x² - 81  =  x²- 9²  ==>  (x+9)(x-9)

x² - 4  =  x²- 2²  ==> (x+2)(x-2)

x²+6x+8  =  (x+2)(x+4)

x²-5x-36  =  (x-9)(x+4)

f(x)  =  [(x+9)(x-9)/(x+2)(x-2)] ⋅ [(x+2)(x+4)/(x-9)(x+4)]

By simplifying (x+9)/(x-2)

So, the value of f(x) is (x+9)/(x-2).

Problem 3 :

[(x²-3x-10)/(x²-x-20)] ⋅ [(x²-2x+4)/(x³+8)]

Solution :

Let f(x)  =  [(x²-3x-10)/(x²-x-20)] ⋅ [(x²-2x+4)/(x³+8)]

x²-3x-10  =  (x-5)(x+2)

x²-x-20  =  (x-5)(x+4)

a³+b³  =  (a+b)(a²-ab+b²)

x³+2³  =  (x+2)(x²-2x+4)

By applying the factors in f(x), we get

 =  [(x-5)(x+2)/(x-5)(x+4)] ⋅ [(x²-2x+4)/(x+2)(x²-2x+4)]

=  1/(x+4)

So, the value of f(x) is 1/(x+4)

Problem 4 :

[(x²-16)/(x^2-3x+2)] ⋅ [(x²-4)/(x³+64)] ⋅ 

[(x²-4x+16)/(x²-2x-8)]

Solution :

Let f(x)  =  [(x²-16)/(x^2-3x+2)] ⋅ [(x²-4)/(x³+64)] ⋅ 

[(x²-4x+16)/(x²-2x-8)]e

x²-16  =  x²-4²  ==>  (x+4)(x-4)

x²-3x+2  =  (x-1)(x-2)

x²-4  =  x²-2²  ==>  (x+2)(x-2)

x³+64  =  x³+4³  ==> (x+4)(x²-4x+16) 

x²-2x-8  =  (x-4)(x+2)

=  [(x+4)(x-4)/(x-1)(x-2)]⋅[(x+2)(x-2)/(x+4)(x²-4x+16)]

⋅[(x²-4x+16)/(x-4)(x+2)]

f(x)  =  1/(x-1)

So, the value of f(x) is 1/(x-1).

Problem 5 :

[(3x²+2x-1)/(x²-x-2)] [(2x²-3x-2)/(3x²+5x-2)]

Solution :

Let f(x)  =  [(3x²+2x-1)/(x²-x-2)]⋅

 [(2x²-3x-2)/(3x²+5x-2)]

(3x²+2x-1)  =  (3x-1) (x+1)

(x²-x-2)  =  (x-2) (x+1)

(2x²-3x-2)  =  (2x+1) (x-2)

(3x²+5x-2) =   (2x-1) (x+2)

By applying the factors in f(x), we get

=  [(3x-1)(x+1)/(x-2) (x+1)]⋅[(2x+1) (x-2)/(2x-1) (x+2)]

=  (2x+1)/(x+2)

So, the value of f(x) is (2x+1)/(x+2).

Problem 6 :

[(2x-1)/(x²+2x+4)] ⋅[(x⁴-8x)/(2x²+5x-3)] ⋅

[(x+3)/(x²-2x)]

Solution :

Let f(x)  =  [(2x-1)/(x²+2x+4)] ⋅[(x⁴-8x)/(2x²+5x-3)] ⋅

[(x+3)/(x²-2x)]

x⁴-8x  =  x(x³-2³)

x⁴-8x  =  x(x-2)(x²+2x+4)

2x²+5x-3  =  (2x-1)(x+3)

x²-2x  =  x(x-2)

By applying the factors in f(x), we get

=  [(2x-1)/(x²+2x+4)]⋅[x(x-2)(x²+2x+4)/(2x-1)(x+3)] ⋅

[(x+3)/x(x-2)]

=  1

So, the value of f(x) is 1.

Problem 7 :

[(a+b)/(a-b)] [(a³-b³)/(a³+b³)]

Solution :

Let f(x)  =  [(a+b)/(a-b)] [(a³-b³)/(a³+b³)]

=  [(a+b)/(a-b)]⋅[(a-b)(a²+ab+b²)/(a+b) (a²-ab+b²)]

=  (a²+ab+b²)/(a²-ab+b²)

So, the value of f(x) is (a²+ab+b²)/(a²-ab+b²).

Problem 8 :

[(x²-9y²)/(3x-3y)] ⋅ [(x²-y²)/(x²+4xy+3y²)]

Solution :

Let f(x)  =  [(x²-9y²)/(3x-3y)] ⋅ [(x²-y²)/(x²+4xy+3y²)]

x²-9y²  =  x²-(3y)²

x²-9y²  =  (x+3y)(x-3y)

3x-3y  =  3(x-y)

x²-y²  =  (x+y)(x-y)

x²+4xy+3y²  =  (x+3y)(x+y)

By applying the factors in f(x), we get

=  [(x+3y)(x-3y)/3(x-y)]⋅[(x+y)(x-y)/(x+3y)(x+y)]

By simplifying, we get

=  (x-3y)/3

So, the value of f(x) is (x-3y)/3.

Problem 9 :

[(x²-4x-12)/(x²-3x-18)] ⋅ [(x²-2x-3)/(x²+3x+2)]

Solution :

Let f(x)  =  [(x²-4x-12)/(x²-3x-18)] 

⋅ [(x²-2x-3)/(x²+3x+2)]

x²-4x-12  =  (x-6)(x+2)

x²-3x-18  =  (x-6)(x+3)

x²-2x-3  =  (x-3)(x+1)

x²+3x+2  =  (x+1)(x+2)

f(x)  =  [(x-6)(x+2)/(x-6)(x+3)]⋅[(x-3)(x+1)/(x+1)(x+2)]

f(x)  =  (x-3)/(x+3)

So, the value of f(x) is (x-3)/(x+3).

Problem 10 :

[(x²-3x-10)/(x²-x-20)]⋅[(x²-4x+16)/(x³+64)]

Solution :

Let f(x)  =  [(x²-3x-10)/(x²-x-20)]⋅[(x²-4x+16)/(x³+64)]

x²-3x-10  =  (x-5)(x+2)

x²-x-20  =  (x-5)(x+4)

x³+4³  =  (x+4)(x²-4x+16)

By applying the factors in f(x), we get

f(x)  =  [(x-5)(x+2)/(x-5)(x+4)]⋅[(x²-4x+16)/(x+4)(x²-4x+16)]

f(x)  =  (x+2)/(x+4)²

So, the value of f(x) is (x+2)/(x+4)².

Problem 11 :

 [(x²-16)/(x-2)] [(x²-4)/(x³+64)]

Solution :

Let f(x)  =  [(x²-16)/(x-2)] [(x²-4)/(x³+64)]

x²-16  =  x²-4²  ==>  (x+4)(x-4)

x²-4  =  x²-2²  ==>  (x+2)(x-2)

x³+64  =  x³+4³  ==>  (x+4)(x²-4x+16)

f(x)  =  [(x+4)(x-4)/(x-2)] [(x+2)(x-2)/(x+4)(x²-4x+16)]

f(x)  =  (x-4)(x-2)/(x²-4x+16)

So, the value of f(x) is (x-4)(x-2)/(x²-4x+16).

Problem 12 :

[(x + 7)/(x²+14x+49)] [(x²+8x+7)/(x+1)]

Solution :

Let f(x)  =  [(x + 7)/(x²+14x+49)] [(x²+8x+7)/(x+1)]

x²+14x+49  =  (x+7)(x+7)

x²+8x+7  =  (x+1)(x+7)

By applying the factors in f(x), we get

f(x)  =  [(x+7)/(x+7)(x+7)] [(x+1)(x+7)/(x+1)]

f(x)  =  1

So, the value of f(x) is 1.

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