# Factor 7x 14 ## Reducing fractions to their lowest terms

### Step  1  :

14 Simplify —— x

#### Equation at the end of step  1  :

14 (7x - ——) - 2 x

### Step  2  :

#### Rewriting the whole as an Equivalent Fraction :

2.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  x  as the denominator :

7x 7x • x 7x = —— = —————— 1 x

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

#### Adding fractions that have a common denominator :

2.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

7x • x - (14) 7x2 - 14 ————————————— = ———————— x x

#### Equation at the end of step  2  :

(7x2 - 14) —————————— - 2 x

### Step  3  :

#### Rewriting the whole as an Equivalent Fraction :

3.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  x  as the denominator :

2 2 • x 2 = — = ————— 1 x

### Step  4  :

#### Pulling out like terms :

4.1     Pull out like factors :

7x2 - 14  =   7 • (x2 - 2)

#### Trying to factor as a Difference of Squares :

4.2      Factoring:  x2 - 2

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2- AB + AB - B2 =
A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 2 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

#### Adding fractions that have a common denominator :

4.3       Adding up the two equivalent fractions

7 • (x2-2) - (2 • x) 7x2 - 2x - 14 ———————————————————— = ————————————— x x

#### Trying to factor by splitting the middle term

4.4     Factoring  7x2 - 2x - 14

The first term is,  7x2  its coefficient is  7 .
The middle term is,  -2x  its coefficient is  -2 .
The last term, "the constant", is  -14

Step-1 : Multiply the coefficient of the first term by the constant   7 • -14 = -98

Step-2 : Find two factors of  -98  whose sum equals the coefficient of the middle term, which is   -2 .

 -98 + 1 = -97 -49 + 2 = -47 -14 + 7 = -7 -7 + 14 = 7 -2 + 49 = 47 -1 + 98 = 97

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

### Final result :

7x2 - 2x - 14 ————————————— x
Sours: https://www.tiger-algebra.com/drill/7x-14/x-2/

## Simplification or other simple results

### Reformatting the input :

(1): "x2"   was replaced by   "x^2".

(7x2 - 7x) - 14

### Step  3  :

#### Pulling out like terms :

3.1     Pull out like factors :

7x2 - 7x - 14  =   7 • (x2 - x - 2)

#### Trying to factor by splitting the middle term

3.2     Factoring  x2 - x - 2

The first term is,  x2  its coefficient is  1 .
The middle term is,  -x  its coefficient is  -1 .
The last term, "the constant", is  -2

Step-1 : Multiply the coefficient of the first term by the constant   1 • -2 = -2

Step-2 : Find two factors of  -2  whose sum equals the coefficient of the middle term, which is   -1 .

 -2 + 1 = -1 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -2  and  1
x2 - 2x + 1x - 2

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-2)
Add up the last 2 terms, pulling out common factors :
1 • (x-2)
Step-5 : Add up the four terms of step 4 :
(x+1)  •  (x-2)
Which is the desired factorization

### Final result :

7 • (x + 1) • (x - 2)
Sours: https://www.tiger-algebra.com/drill/7x2-7x-14/  ### Most Used Actions

 \mathrm{simplify} \mathrm{solve\:for} \mathrm{inverse} \mathrm{tangent} \mathrm{line}
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