Solve by Factoring (x^2-6x)^2-35(x^2-6x)-200=0

${(x^{2} - 6 x)}^{2} - 35 (x^{2} - 6 x) - 200 = 0$

Step 1

Let $u = x^{2} - 6 x$ . Substitute $u$ for all occurrences of $x^{2} - 6 x$ .

$u^{2} - 35 u - 200 = 0$

Step 2

Factor $u^{2} - 35 u - 200$ 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 $- 200$ and whose sum is $- 35$ .

$- 40, 5$

Write the factored form using these integers.

$(u - 40) (u + 5) = 0$

Step 3

Replace all occurrences of $u$ with $x^{2} - 6 x$ .

$(x^{2} - 6 x - 40) (x^{2} - 6 x + 5) = 0$

Step 4

Factor $x^{2} - 6 x - 40$ 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 $- 40$ and whose sum is $- 6$ .

$- 10, 4$

Write the factored form using these integers.

$(x - 10) (x + 4) (x^{2} - 6 x + 5) = 0$

Step 5

Factor.

Factor $x^{2} - 6 x + 5$ 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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$(x - 10) (x + 4) ((x - 5) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x - 10) (x + 4) (x - 5) (x - 1) = 0$

Step 6

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

$x - 10 = 0$

$x + 4 = 0$

$x - 5 = 0$

$x - 1 = 0$

Step 7

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Step 8

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 9

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 10

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 11

The final solution is all the values that make $(x - 10) (x + 4) (x - 5) (x - 1) = 0$ true.

$x = 10, - 4, 5, 1$

Solve by Factoring (x^2-6x)^2-35(x^2-6x)-200=0

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