Precalculus Examples

$f (x) = 2 x^{3} + 7 x^{2} - 5 x - 0$

Step 1

Regroup terms.

$f (x) = 2 x^{3} + 7 x^{2} - 5 x - 0$

Step 2

Factor $x^{2}$ out of $2 x^{3} + 7 x^{2}$ .

Tap for more steps...

Step 2.1

Factor $x^{2}$ out of $2 x^{3}$ .

$f (x) = x^{2} (2 x) + 7 x^{2} - 5 x - 0$

Step 2.2

Factor $x^{2}$ out of $7 x^{2}$ .

$f (x) = x^{2} (2 x) + x^{2} \cdot 7 - 5 x - 0$

Step 2.3

Factor $x^{2}$ out of $x^{2} (2 x) + x^{2} \cdot 7$ .

$f (x) = x^{2} (2 x + 7) - 5 x - 0$

Step 3

Multiply $- 1$ by $0$ .

$f (x) = x^{2} (2 x + 7) - 5 x + 0$

Step 4

Add $- 5 x$ and $0$ .

$f (x) = x^{2} (2 x + 7) - 5 x$

Step 5

Factor $x$ out of $x^{2} (2 x + 7) - 5 x$ .

Tap for more steps...

Step 5.1

Factor $x$ out of $x^{2} (2 x + 7)$ .

$f (x) = x (x (2 x + 7)) - 5 x$

Step 5.2

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

$f (x) = x (x (2 x + 7)) + x \cdot - 5$

Step 5.3

Factor $x$ out of $x (x (2 x + 7)) + x \cdot - 5$ .

$f (x) = x (x (2 x + 7) - 5)$

Step 6

Apply the distributive property.

$f (x) = x (x (2 x) + x \cdot 7 - 5)$

Step 7

Rewrite using the commutative property of multiplication.

$f (x) = x (2 x \cdot x + x \cdot 7 - 5)$

Step 8

Move $7$ to the left of $x$ .

$f (x) = x (2 x \cdot x + 7 \cdot x - 5)$

Step 9

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 9.1

Move $x$ .

$f (x) = x (2 (x \cdot x) + 7 \cdot x - 5)$

Step 9.2

Multiply $x$ by $x$ .

$f (x) = x (2 x^{2} + 7 \cdot x - 5)$

$f (x) = x (2 x^{2} + 7 x - 5)$