Linear Algebra Examples

$4 x^{2} + 3 x - 1 = 0$

Step 1

Factor by grouping.

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Step 1.1

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 $a \cdot c = 4 \cdot - 1 = - 4$ and whose sum is $b = 3$ .

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Step 1.1.1

Factor $3$ out of $3 x$ .

$4 x^{2} + 3 (x) - 1 = 0$

Step 1.1.2

Rewrite $3$ as $- 1$ plus $4$

$4 x^{2} + (- 1 + 4) x - 1 = 0$

Step 1.1.3

Apply the distributive property.

$4 x^{2} - 1 x + 4 x - 1 = 0$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$(4 x^{2} - 1 x) + 4 x - 1 = 0$

Step 1.2.2

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

$x (4 x - 1) + 1 (4 x - 1) = 0$

Step 1.3

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

$(4 x - 1) (x + 1) = 0$

Step 2

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

$4 x - 1 = 0$

$x + 1 = 0$

Step 3

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

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Step 3.1

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

$4 x - 1 = 0$

Step 3.2

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

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Step 3.2.1

Add $1$ to both sides of the equation.

$4 x = 1$

Step 3.2.2

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

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Step 3.2.2.1

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

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

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $4$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

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

Step 3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 4

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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

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

Step 6

The domain is the set of all valid $x$ values.

${x | x = \frac{1}{4}, - 1}$

Step 7