When dividing a^2+a-2/a^2+5a+6 divided by a-1/a, what are the term(s) that can be cancelled

Oops. Please don't omit parenthesis when showing polynomial fractions.

I answered the question you asked.

First don't forget that a can't be zero.

Maybe the author of the question is asking whether I notice that there are two _terms_, a and 5a, which should be _combined_ to form 6a before beginning to do polynomial long division.

Maybe I should answer that I can multiply numerator and denominator by a^2.

Maybe the author meant to write "factors" rather than "terms" and expects to perform the division by factoring numerator and denominator and _cancelling_ identical _factors_ which appear in numerator and denominator. That looks harder than long division, and the easiest way to discover common factors might be to do the division.

It turns out there are no common factors in divisor and dividend.

(a^2+a-2/a^2+5a+6) / (a-1/a)

quotient: (a+6)

remainder: (7+6/a-2/a^2)

divisor: (a-1/a)

Step-by-step explanation:

This isn't a clear question: you _combine like_ terms, not _cancel_ terms. Terms get removed when they add to zero, while factors in numerator and denominator get _cancelled_ when together they divide to give one, not zero.

First note that a can't be zero.

Second simplify the polynomials by combining like terms, and reorder the terms in descending powers of a, and replace 1/a^n by a^-n.

dividend

a^2+a-2/a^2+5a+6 becomes

a^2+6a+6-2a^-2

divisor a-1/a becomes a-a^-1

Then clear denominators in the terms by multiplying dividend and divisor by a^2, so the problem becomes

(a^4+6a^3+6a^2-2) / (a^3-a)

Finally, first term of quotient is a.

Subtract a^4-a^2 from dividend to get first remainder

6a^3+7a^2-2

Second term of quotient is 6.

Subtract 6a^3-6a from first remainder to get second remainder

7a^2+6a-2

Since the divisor is a^3-a, we are done.

(a^4+6a^3+6a^2-2) / (a^3-a) =

quotient (a+6)

remainder (7a^2+6a-2)

divisor (a^3-a)

The original problem has the same quotient (a+6) but the remainder has to be divided by a^2: 7+6/a-2/a^2

(a^2+a-2/a^2+5a+6) / (a-1/a)

quotient: (a+6)

remainder: (7+6/a-2/a^2)

divisor: (a-1/a)

Check: Quotient × divisor + remainder = dividend

(a+6)(a-1/a) + (7+6/a-2/a^2)

= (a^2 + 6a - 1 - 6/a) + (7+6/a-2/a^2)

= a^2 + 6a + 6 - 2/a^2

which is indeed equal to the original dividend a^2+a-2/a^2+5a+6

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.