Solve by Factoring 2x^-2-13x^-1+20=0

$2 x^{- 2} - 13 x^{- 1} + 20 = 0$

Step 1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 (\frac{1}{x^{2}}) - 13 x^{- 1} + 20 = 0$

Step 2

Combine $2$ and $\frac{1}{x^{2}}$ .

$\frac{2}{x^{2}} - 13 x^{- 1} + 20 = 0$

Step 3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{2}{x^{2}} - 13 \frac{1}{x} + 20 = 0$

Step 4

Combine $- 13$ and $\frac{1}{x}$ .

$\frac{2}{x^{2}} + \frac{- 13}{x} + 20 = 0$

Step 5

Move the negative in front of the fraction.

$\frac{2}{x^{2}} - \frac{13}{x} + 20 = 0$

Step 6

Reorder terms.

$- \frac{13}{x} + \frac{2}{x^{2}} + 20 = 0$

Step 7

To write $- \frac{13}{x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$- \frac{13}{x} \cdot \frac{x}{x} + \frac{2}{x^{2}} + 20 = 0$

Step 8

Write each expression with a common denominator of $x^{2}$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{13}{x}$ by $\frac{x}{x}$ .

$- \frac{13 x}{x \cdot x} + \frac{2}{x^{2}} + 20 = 0$

Raise $x$ to the power of $1$ .

$- \frac{13 x}{x \cdot x} + \frac{2}{x^{2}} + 20 = 0$

Raise $x$ to the power of $1$ .

$- \frac{13 x}{x \cdot x} + \frac{2}{x^{2}} + 20 = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{13 x}{x^{1 + 1}} + \frac{2}{x^{2}} + 20 = 0$

Add $1$ and $1$ .

$- \frac{13 x}{x^{2}} + \frac{2}{x^{2}} + 20 = 0$

Step 9

Combine the numerators over the common denominator.

$\frac{- 13 x + 2}{x^{2}} + 20 = 0$

Step 10

To write $20$ as a fraction with a common denominator, multiply by $\frac{x^{2}}{x^{2}}$ .

$\frac{- 13 x + 2}{x^{2}} + \frac{20 x^{2}}{x^{2}} = 0$

Step 11

Combine the numerators over the common denominator.

$\frac{- 13 x + 2 + 20 x^{2}}{x^{2}} = 0$

Step 12

Factor by grouping.

Reorder terms.

$\frac{20 x^{2} - 13 x + 2}{x^{2}} = 0$

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 $a \cdot c = 20 \cdot 2 = 40$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$\frac{20 x^{2} - 13 x + 2}{x^{2}} = 0$

Rewrite $- 13$ as $- 5$ plus $- 8$

$\frac{20 x^{2} + (- 5 - 8) x + 2}{x^{2}} = 0$

Apply the distributive property.

$\frac{20 x^{2} - 5 x - 8 x + 2}{x^{2}} = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(20 x^{2} - 5 x) - 8 x + 2}{x^{2}} = 0$

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

$\frac{5 x (4 x - 1) - 2 (4 x - 1)}{x^{2}} = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$\frac{(4 x - 1) (5 x - 2)}{x^{2}} = 0$

Step 13

Set the numerator equal to zero.

$(4 x - 1) (5 x - 2) = 0$

Step 14

Solve the equation for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 x - 1 = 0$

$5 x - 2 = 0$

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Set $5 x - 2$ equal to $0$ and solve for $x$ .

Set $5 x - 2$ equal to $0$ .

$5 x - 2 = 0$

Solve $5 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$5 x = 2$

Divide each term in $5 x = 2$ by $5$ and simplify.

Divide each term in $5 x = 2$ by $5$ .

$\frac{5 x}{5} = \frac{2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{2}{5}$

Divide $x$ by $1$ .

$x = \frac{2}{5}$

The final solution is all the values that make $(4 x - 1) (5 x - 2) = 0$ true.

$x = \frac{1}{4}, \frac{2}{5}$

Solve by Factoring 2x^-2-13x^-1+20=0

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