Solve by Factoring (x-4)^2-(x-4)-20=0

${(x - 4)}^{2} - (x - 4) - 20 = 0$

Step 1

Let $u = x - 4$ . Substitute $u$ for all occurrences of $x - 4$ .

$u^{2} - u - 20 = 0$

Step 2

Factor $u^{2} - u - 20$ 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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$(u - 5) (u + 4) = 0$

Step 3

Replace all occurrences of $u$ with $x - 4$ .

$(x - 4 - 5) (x - 4 + 4) = 0$

Step 4

Simplify.

Subtract $5$ from $- 4$ .

$(x - 9) (x - 4 + 4) = 0$

Combine the opposite terms in $x - 4 + 4$ .

Add $- 4$ and $4$ .

$(x - 9) (x + 0) = 0$

Add $x$ and $0$ .

$(x - 9) x = 0$

Step 5

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

$x - 9 = 0$

$x = 0$

Step 6

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Step 7

Set $x$ equal to $0$ .

$x = 0$

Step 8

The final solution is all the values that make $(x - 9) x = 0$ true.

$x = 9, 0$

Solve by Factoring (x-4)^2-(x-4)-20=0

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