Find the Roots (Zeros) 2x^4-7x^3+6x^2=0

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

Step 1

Factor the left side of the equation.

Factor $x^{2}$ out of $2 x^{4} - 7 x^{3} + 6 x^{2}$ .

Factor $x^{2}$ out of $2 x^{4}$ .

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

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

$x^{2} (2 x^{2}) + x^{2} (- 7 x) + 6 x^{2} = 0$

Factor $x^{2}$ out of $6 x^{2}$ .

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

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

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

Factor $x^{2}$ out of $x^{2} (2 x^{2} - 7 x) + x^{2} \cdot 6$ .

$x^{2} (2 x^{2} - 7 x + 6) = 0$

Factor.

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 $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$x^{2} (2 x^{2} - 7 x + 6) = 0$

Rewrite $- 7$ as $- 3$ plus $- 4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 2

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

$x^{2} = 0$

$2 x - 3 = 0$

$x - 2 = 0$

Step 3

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 4

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 5

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 6

The final solution is all the values that make $x^{2} (2 x - 3) (x - 2) = 0$ true.

$x = 0, \frac{3}{2}, 2$

Step 7

Find the Roots (Zeros) 2x^4-7x^3+6x^2=0

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