Factoring With Fractions

Factoring is the process of finding the factor of the given function. It is otherwise called as divide the given function as two separate part. We can factor the function more than two times. Fractions are also used in factoring process. Fraction means it has some numerator and denominator values. Both the numerator and denominator values are different. Now in this article, we see about ac factoring with fractions and their example problems.

Example problems for Ac factoring with fractions

Ac factoring with fractions example problem 1:

Factorize the given polynomial fraction equation `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

Solution:

Given equation is `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 6 on both the sides, we get

x2 - 7x - 120 = 0

Factorize the above equation, we get

x2 - 15x + 8x - 120 = 0

Grouping the first two terms and next two terms, we get

(x2 - 15x) + (8x - 120) = 0

x (x - 15) + 8 (x - 15) = 0

(x - 15) (x + 8) = 0

The factors are (x - 15) and (x + 8)

Answer:

The final answer is (x - 15) and (x + 8)

Ac factoring with fractions example problem 2:

Factorize the given polynomial fraction equation `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

Solution:

Given equation is `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 9 on both the sides, we get

x2 - 12x + 11 = 0

Factorize the above equation, we get

x2 - 11x - x + 11 = 0

Grouping the first two terms and next two terms, we get

(x2 - 11x) + (- x + 11) = 0

x (x - 11) - 1(x - 11) = 0

(x - 11) (x - 1) = 0

The factors are (x - 11) and (x - 1)

Answer:

The final answer is (x - 11) and (x - 1)

Ac factoring with fractions example problem 3:

Factorize the given polynomial fraction equation `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Solution:

Given equation is `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Multiply the given equation by 2 on both the sides, we get

x2 - 11x + 28 = 0

Factorize the above equation, we get

x2 - 7x - 4x + 28 = 0

Grouping the first two terms and next two terms, we get

(x2 - 7x) + (- 4x + 28) = 0

x (x - 7) - 4 (x - 7) = 0

(x - 4) (x - 7) = 0

The factors are (x - 4) and (x - 7)

My forthcoming post is on Rules for Dividing Integers and class 11 cbse books will give you more understanding about Algebra.

Answer:

The final answer is (x - 4) and (x - 7)