Find a,b,c x(x^2+3x-10)=0

$x (x^{2} + 3 x - 10) = 0$

Simplify the left side.

Simplify $x (x^{2} + 3 x - 10)$ .

Apply the distributive property.

$x \cdot x^{2} + x (3 x) + x \cdot - 10 = 0$

Simplify.

Multiply $x$ by $x^{2}$ by adding the exponents.

Multiply $x$ by $x^{2}$ .

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

$x^{1} x^{2} + x (3 x) + x \cdot - 10 = 0$

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

$x^{1 + 2} + x (3 x) + x \cdot - 10 = 0$

Add $1$ and $2$ .

$x^{3} + x (3 x) + x \cdot - 10 = 0$

Rewrite using the commutative property of multiplication.

$x^{3} + 3 x \cdot x + x \cdot - 10 = 0$

Move $- 10$ to the left of $x$ .

$x^{3} + 3 x \cdot x - 10 \cdot x = 0$

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

Move $x$ .

$x^{3} + 3 (x \cdot x) - 10 \cdot x = 0$

Multiply $x$ by $x$ .

$x^{3} + 3 x^{2} - 10 \cdot x = 0$

$x^{3} + 3 x^{2} - 10 x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 3 x^{2} - 10 x$ .

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

$x \cdot x^{2} + 3 x^{2} - 10 x = 0$

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

$x \cdot x^{2} + x (3 x) - 10 x = 0$

Factor $x$ out of $- 10 x$ .

$x \cdot x^{2} + x (3 x) + x \cdot - 10 = 0$

Factor $x$ out of $x \cdot x^{2} + x (3 x)$ .

$x (x^{2} + 3 x) + x \cdot - 10 = 0$

Factor $x$ out of $x (x^{2} + 3 x) + x \cdot - 10$ .

$x (x^{2} + 3 x - 10) = 0$

Factor.

Factor $x^{2} + 3 x - 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

$x ((x - 2) (x + 5)) = 0$

Remove unnecessary parentheses.

$x (x - 2) (x + 5) = 0$

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

$x = 0$

$x - 2 = 0$

$x + 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 2$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = 0, 2, - 5$

Find a,b,c x(x^2+3x-10)=0

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