Simplify (2x^2-11x-21)/(5x^2+16x+3)*(5x+1)/(5x-35)

$\frac{2 x^{2} - 11 x - 21}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

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 $a \cdot c = 2 \cdot - 21 = - 42$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$\frac{2 x^{2} - 11 (x) - 21}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Rewrite $- 11$ as $3$ plus $- 14$

$\frac{2 x^{2} + (3 - 14) x - 21}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Apply the distributive property.

$\frac{2 x^{2} + 3 x - 14 x - 21}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{2} + 3 x) - 14 x - 21}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

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

$\frac{x (2 x + 3) - 7 (2 x + 3)}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$\frac{(2 x + 3) (x - 7)}{5 x^{2} + 16 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Step 2

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 $a \cdot c = 5 \cdot 3 = 15$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$\frac{(2 x + 3) (x - 7)}{5 x^{2} + 16 (x) + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Rewrite $16$ as $1$ plus $15$

$\frac{(2 x + 3) (x - 7)}{5 x^{2} + (1 + 15) x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Apply the distributive property.

$\frac{(2 x + 3) (x - 7)}{5 x^{2} + 1 x + 15 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Multiply $x$ by $1$ .

$\frac{(2 x + 3) (x - 7)}{5 x^{2} + x + 15 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x + 3) (x - 7)}{(5 x^{2} + x) + 15 x + 3} \cdot \frac{5 x + 1}{5 x - 35}$

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

$\frac{(2 x + 3) (x - 7)}{x (5 x + 1) + 3 (5 x + 1)} \cdot \frac{5 x + 1}{5 x - 35}$

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

$\frac{(2 x + 3) (x - 7)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5 x - 35}$

Step 3

Factor $5$ out of $5 x - 35$ .

Factor $5$ out of $5 x$ .

$\frac{(2 x + 3) (x - 7)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5 (x) - 35}$

Factor $5$ out of $- 35$ .

$\frac{(2 x + 3) (x - 7)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5 x + 5 \cdot - 7}$

Factor $5$ out of $5 x + 5 \cdot - 7$ .

$\frac{(2 x + 3) (x - 7)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5 (x - 7)}$

Step 4

Cancel the common factor of $x - 7$ .

Factor $x - 7$ out of $(2 x + 3) (x - 7)$ .

$\frac{(x - 7) (2 x + 3)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5 (x - 7)}$

Factor $x - 7$ out of $5 (x - 7)$ .

$\frac{(x - 7) (2 x + 3)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{(x - 7) \cdot 5}$

Cancel the common factor.

$\frac{(x - 7) (2 x + 3)}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{(x - 7) \cdot 5}$

Rewrite the expression.

$\frac{2 x + 3}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5}$

Step 5

Cancel the common factor of $5 x + 1$ .

Cancel the common factor.

$\frac{2 x + 3}{(5 x + 1) (x + 3)} \cdot \frac{5 x + 1}{5}$

Rewrite the expression.

$\frac{2 x + 3}{x + 3} \cdot \frac{1}{5}$

Step 6

Multiply $\frac{2 x + 3}{x + 3}$ by $\frac{1}{5}$ .

$\frac{2 x + 3}{(x + 3) \cdot 5}$

Step 7

Move $5$ to the left of $x + 3$ .

$\frac{2 x + 3}{5 (x + 3)}$

Simplify (2x^2-11x-21)/(5x^2+16x+3)*(5x+1)/(5x-35)

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