(8x^3-27)/(4x^2-9)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(8x^3- 27)/(4X^2 - 9)

To simplify the fraction,

We know that:

(a^3 - b^3) = (a-b) ( a^2 + ab + b^2)

(a^2 - b^2) = (a-b) *(a+b)

LET US factor:

==> (2x-3)(4x^2 + 6x + 9) / (2x-3)(2x+3)

Now reduce similars:

==> (4x^2 + 6x + 9)/(2x+3)

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To evaluate the expression we'll use factorization. We notice that the numerator is a difference of cubes:

8x^3-27 = (2x)^3 - (3)^3

We'll apply the formula:

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

We'll put a = 2x and b = 3

(2x)^3 - (3)^3 = (2x-3)(4x^2 + 6x + 9)

We also notice that the denominator is a difference of squares:

4x^2-9 = (2x)^2 - 3^2

We'll apply the formula:

a^2 - b^2 = (a-b)(a+b)

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

We'll substitute the differences by their products:

 [(8x^3-27)/(4x^2-9)] = (2x-3)(4x^2 + 6x + 9)/(2x-3)(2x+3)]

We'll simplify by the common factor (2x-3):

 [(8x^3-27)/(4x^2-9)] =  [(4x^2 + 6x + 9)/(2x+3)]

We can also combine the terms 6x + 9 and factorize them by 3;

[(8x^3-27)/(4x^2-9)] = 4x^2/(2x+3) + (6x + 9)/(2x+3)

[(8x^3-27)/(4x^2-9)] = 4x^2/(2x+3) + 3(2x + 3)/(2x+3)

We'll simplify the last ratio by (2x+3) and we'll get:

[(8x^3-27)/(4x^2-9)] = [4x^2/(2x+3)] + 3

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

We notice that the given expression is in the form:

(a^3 - b^3)/(a^2 - b^2)

Where:

a = 2x and

b  = 3

We factorise both the numerator and denominator of the given expression using the formulas:

a^3 - b^3 = (a - b)(a^2 +ab + b^2)

a^2 - b^2 = (a - b)(a + b)

Thus:

(8x^3 - 27)/(4x^2 - 9)

= [(2x - 3)(4x^2 + 6x + 9]/[(2x - 3)(2x + 3)]

= (4x^2 + 6x + 9/(2x + 3)

This can be further simplified as:

= (4x^2 + 6x)/(2x + 3) + 9/(2x + 3)

= 2x + 9/(2x + 3)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

(8x^3-27)/(4x^2-9).

We can directly divide (8x^2-27) by (4x^2-9).

Or we  can factorise both numerator and deniminator and then reduce by the common factors.

8x^3-27 = (2x)^3-3^3 = (2x-3)(4x^2+6x+9), as a^3 -b^3 = (a-b)(a^2+ab+b^2)

4x^-9 = (2x)^2 - 3^ = (2x-3)(2x+3) , as a^2-b^2 = (a-b)(a+b).

Therefore the numerator in the rational eexpression is reduced to (4x^2+6x+9) and denominator to 2x+3 after dividing by the common factor 2x-3.

So (8x^3-27)/(4x^2-9) = (4x^2+6x+9)/(2x+3)

If we want further division in the for  full quotient and  and a lowest irreducible rational fraction, we go as below:

Numerator(4x^2+6x+9) = (2x+3)^2 - 6x.

Therefore (8x^3-27)/(4x^2-9) =  (2x+3) - 6x/(2x+3).

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