Pre-Algebra Examples

$x^{4} - x^{3} - 7 x^{2} + x + 6 = 0$

Step 1

Move all terms not containing $4$ to the right side of the equation.

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Step 1.1

Subtract $x^{4}$ from both sides of the equation.

$- x^{3} - 7 x^{2} + x + 6 = - x^{4}$

Step 1.2

Add $x^{3}$ to both sides of the equation.

$- 7 x^{2} + x + 6 = - x^{4} + x^{3}$

Step 1.3

Add $7 x^{2}$ to both sides of the equation.

$x + 6 = - x^{4} + x^{3} + 7 x^{2}$

Step 1.4

Subtract $x$ from both sides of the equation.

$6 = - x^{4} + x^{3} + 7 x^{2} - x$

Step 1.5

Subtract $6$ from both sides of the equation.

$0 = - x^{4} + x^{3} + 7 x^{2} - x - 6$

Step 2

Rewrite the equation as $- x^{4} + x^{3} + 7 x^{2} - x - 6 = 0$ .

$- x^{4} + x^{3} + 7 x^{2} - x - 6 = 0$

Step 3

Divide each term in $- x^{4} + x^{3} + 7 x^{2} - x - 6 = 0$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- x^{4} + x^{3} + 7 x^{2} - x - 6 = 0$ by $- 1$ .

$\frac{- x^{4}}{- 1} + \frac{x^{3}}{- 1} + \frac{7 x^{2}}{- 1} + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Dividing two negative values results in a positive value.

$\frac{x^{4}}{1} + \frac{x^{3}}{- 1} + \frac{7 x^{2}}{- 1} + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.2

Divide $x^{4}$ by $1$ .

$x^{4} + \frac{x^{3}}{- 1} + \frac{7 x^{2}}{- 1} + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.3

Move the negative one from the denominator of $\frac{x^{3}}{- 1}$ .

$x^{4} - 1 \cdot x^{3} + \frac{7 x^{2}}{- 1} + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.4

Rewrite $- 1 \cdot x^{3}$ as $- x^{3}$ .

$x^{4} - x^{3} + \frac{7 x^{2}}{- 1} + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.5

Move the negative one from the denominator of $\frac{7 x^{2}}{- 1}$ .

$x^{4} - x^{3} - 1 \cdot (7 x^{2}) + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.6

Rewrite $- 1 \cdot (7 x^{2})$ as $- (7 x^{2})$ .

$x^{4} - x^{3} - (7 x^{2}) + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.7

Multiply $7$ by $- 1$ .

$x^{4} - x^{3} - 7 x^{2} + \frac{- x}{- 1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.8

Dividing two negative values results in a positive value.

$x^{4} - x^{3} - 7 x^{2} + \frac{x}{1} + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.9

Divide $x$ by $1$ .

$x^{4} - x^{3} - 7 x^{2} + x + \frac{- 6}{- 1} = \frac{0}{- 1}$

Step 3.2.1.10

Divide $- 6$ by $- 1$ .

$x^{4} - x^{3} - 7 x^{2} + x + 6 = \frac{0}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide $0$ by $- 1$ .

$x^{4} - x^{3} - 7 x^{2} + x + 6 = 0$