Simplify (9k^2-33k+10)/(18k^3-200k)

$\frac{9 k^{2} - 33 k + 10}{18 k^{3} - 200 k}$

Step 1

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot 10 = 90$ and whose sum is $b = - 33$ .

Factor $- 33$ out of $- 33 k$ .

$\frac{9 k^{2} - 33 (k) + 10}{18 k^{3} - 200 k}$

Rewrite $- 33$ as $- 3$ plus $- 30$

$\frac{9 k^{2} + (- 3 - 30) k + 10}{18 k^{3} - 200 k}$

Apply the distributive property.

$\frac{9 k^{2} - 3 k - 30 k + 10}{18 k^{3} - 200 k}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(9 k^{2} - 3 k) - 30 k + 10}{18 k^{3} - 200 k}$

Factor out the greatest common factor (GCF) from each group.

$\frac{3 k (3 k - 1) - 10 (3 k - 1)}{18 k^{3} - 200 k}$

Factor the polynomial by factoring out the greatest common factor, $3 k - 1$ .

$\frac{(3 k - 1) (3 k - 10)}{18 k^{3} - 200 k}$

Step 2

Simplify the denominator.

Factor $2 k$ out of $18 k^{3} - 200 k$ .

Factor $2 k$ out of $18 k^{3}$ .

$\frac{(3 k - 1) (3 k - 10)}{2 k (9 k^{2}) - 200 k}$

Factor $2 k$ out of $- 200 k$ .

$\frac{(3 k - 1) (3 k - 10)}{2 k (9 k^{2}) + 2 k (- 100)}$

Factor $2 k$ out of $2 k (9 k^{2}) + 2 k (- 100)$ .

$\frac{(3 k - 1) (3 k - 10)}{2 k (9 k^{2} - 100)}$

Rewrite $9 k^{2}$ as ${(3 k)}^{2}$ .

$\frac{(3 k - 1) (3 k - 10)}{2 k ({(3 k)}^{2} - 100)}$

Rewrite $100$ as $10^{2}$ .

$\frac{(3 k - 1) (3 k - 10)}{2 k ({(3 k)}^{2} - 10^{2})}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 k$ and $b = 10$ .

$\frac{(3 k - 1) (3 k - 10)}{2 k (3 k + 10) (3 k - 10)}$

Step 3

Cancel the common factor of $3 k - 10$ .

Cancel the common factor.

$\frac{(3 k - 1) (3 k - 10)}{2 k (3 k + 10) (3 k - 10)}$

Rewrite the expression.

$\frac{3 k - 1}{2 k (3 k + 10)}$

Simplify (9k^2-33k+10)/(18k^3-200k)

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