Solve by Factoring k(k-4)=-3

$k (k - 4) = - 3$

Step 1

Add $3$ to both sides of the equation.

$k (k - 4) + 3 = 0$

Step 2

Simplify each term.

Apply the distributive property.

$k \cdot k + k \cdot - 4 + 3 = 0$

Multiply $k$ by $k$ .

$k^{2} + k \cdot - 4 + 3 = 0$

Move $- 4$ to the left of $k$ .

$k^{2} - 4 k + 3 = 0$

Step 3

Factor $k^{2} - 4 k + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$(k - 3) (k - 1) = 0$

Step 4

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

$k - 3 = 0$

$k - 1 = 0$

Step 5

Set $k - 3$ equal to $0$ and solve for $k$ .

Set $k - 3$ equal to $0$ .

$k - 3 = 0$

Add $3$ to both sides of the equation.

$k = 3$

Step 6

Set $k - 1$ equal to $0$ and solve for $k$ .

Set $k - 1$ equal to $0$ .

$k - 1 = 0$

Add $1$ to both sides of the equation.

$k = 1$

Step 7

The final solution is all the values that make $(k - 3) (k - 1) = 0$ true.

$k = 3, 1$

Solve by Factoring k(k-4)=-3

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