Solve 3x²+6xy+y³ / x-y

Question

ScanSolve AI Tutor · Accepted Answer

To "solve" this expression means to simplify it, typically by performing the division. We will use polynomial long division to divide the numerator $3x^2+6xy+y^3$ by the denominator $x-y$ .

Set up the polynomial long division.
We are dividing $3x^2+6xy+y^3$ by $x-y$ .
Divide the leading term of the numerator ( $3x^2$ ) by the leading term of the denominator ( $x$ ).
$\frac{3x^2}{x} = 3x$
Multiply the result ( $3x$ ) by the entire denominator ( $x-y$ ):
$3x(x-y) = 3x^2 - 3xy$
Subtract this product from the numerator:
$(3x^2 + 6xy + y^3) - (3x^2 - 3xy) = 3x^2 + 6xy + y^3 - 3x^2 + 3xy = 9xy + y^3$
Divide the leading term of the new polynomial ( $9xy$ ) by the leading term of the denominator ( $x$ ).
$\frac{9xy}{x} = 9y$
Multiply the result ( $9y$ ) by the entire denominator ( $x-y$ ):
$9y(x-y) = 9xy - 9y^2$
Subtract this product from the current polynomial ( $9xy + y^3$ ):
$(9xy + y^3) - (9xy - 9y^2) = 9xy + y^3 - 9xy + 9y^2 = y^3 + 9y^2$
The degree of the remaining polynomial ( $y^3+9y^2$ ) with respect to $x$ (which is 0) is less than the degree of the divisor ( $x-y$ ) with respect to $x$ (which is 1).
Therefore, $y^3+9y^2$ is the remainder.

The quotient is $3x+9y$ and the remainder is $y^3+9y^2$ .
So, the expression can be written as the quotient plus the remainder divided by the divisor.
$\frac{3x^2+6xy+y^3}{x-y} = 3x+9y + \frac{y^3+9y^2}{x-y}$
The simplified expression is $\boxed{3x+9y + \frac{y^3+9y^2}{x-y}}$ .

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Solve 3x²+6xy+y³ / x-y

Answer

Step-by-step solution

Set up the polynomial long division.

Divide the leading term of the numerator ( $3x^2$ ) by the leading term of the denominator ( $x$ ).

Divide the leading term of the new polynomial ( $9xy$ ) by the leading term of the denominator ( $x$ ).

The degree of the remaining polynomial ( $y^3+9y^2$ ) with respect to $x$ (which is 0) is less than the degree of the divisor ( $x-y$ ) with respect to $x$ (which is 1).

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Solve 3x²+6xy+y³ / x-y

Answer

Step-by-step solution

Set up the polynomial long division.

Divide the leading term of the numerator (3x23x^23x2) by the leading term of the denominator (xxx).

Divide the leading term of the new polynomial (9xy9xy9xy) by the leading term of the denominator (xxx).

The degree of the remaining polynomial (y3+9y2y^3+9y^2y3+9y2) with respect to xxx (which is 0) is less than the degree of the divisor (x−yx-yx−y) with respect to xxx (which is 1).

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Still stuck on this one?

More Mathematics Questions

Divide the leading term of the numerator ( $3x^2$ ) by the leading term of the denominator ( $x$ ).

Divide the leading term of the new polynomial ( $9xy$ ) by the leading term of the denominator ( $x$ ).

The degree of the remaining polynomial ( $y^3+9y^2$ ) with respect to $x$ (which is 0) is less than the degree of the divisor ( $x-y$ ) with respect to $x$ (which is 1).