Division (x^2+5x^2+6)+(1/x^2+3)

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kingattaskus12's profile pic

kingattaskus12 | (Level 3) Adjunct Educator

Posted on

Remove the parentheses that are not needed from the expression.

`x^2 + 5x^2 + 6 + 1/(x^2 + 3)`


Since `x^2`  and `5x^2` are like terms, add `5x^2` to `x^2`  to get `6x^2`

`6x^2 + 6 + 1/(x^2 + 3)`


Multiply each term by a factor of `1` that will equate all the denominators. In this case, all terms need a denominator of `(x^2 + 3)`

`6x^2 * (x^2 + 3)/(x^2 + 3) + 6 * (x^2 + 3)/(x^2 + 3) + 1/(x^2 + 3)`


Multiply 6 by each term inside the parentheses.

`(6x^4 + 18x^2)/(x^2 + 3) + (6x^2 + 18)/(x^2 + 3) + 1/(x^2 + 3)`


The numerators of expressions that have equal denominators can be combined. In this case, `((6x^4 + 18x^2))/((x^2 + 3))`  and `((6x^2 + 18))/((x^2 + 3))`  have the same denominator of `(x^2 + 3)`  so the numerators can be combined.

`((6x^4 + 18x^2) + (6x^2 + 18) + 1)/(x^2 + 3)`


Combine all similar expressions in the polynomial.

`(6x^4 + 24x^2 + 19)/(x^2 + 3)`  

sid-sarfraz's profile pic

sid-sarfraz | Student, Graduate | (Level 2) Salutatorian

Posted on

QUESTION:-

Division (x^2+5x^2+6)+(1/x^2+3)

SOLUTION:-

The following equation is supposed to be solved;

`(x^2+5x^2+6)+(1/(x^2+3))`

Taking LCM:-

`(6x^2*(x^2+3)/(x^2+3)+6*(x^2+3)/(x^2+3)+1/(x^2+3))` 

`((6x^2*(x^2+3)+6*(x^2+3)+1)/(x^2+3))`

Now take LCM for the whole equation;

`(6x^4+18x^2+6x^2+18+1)/(x^2+3)`  

`(6x^4+24x^2+19)/(x^2+3)`

Hence Solved

aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

`x^2+5x^2+6+(1/(x^2+3))`

`=6x^2+6+1/(x^2+3)`

`=6(x^2+1)+1/(x^2+3)`

`=(6(x^2+1)(x^2+3)+1)/(x^2+3)`

`=(6(x^4+x^2+3x^2+3)+1)/(x^2+3)`

`=(6x^4+24x^2+18+1)/(x^2+3)`

`=(6x^4+24x^2+19)/(x^2+3)`

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