Find the Roots (Zeros) f(x)=x^2+x-42

$f (x) = x^{2} + x - 42$

Step 1

Set $x^{2} + x - 42$ equal to $0$ .

$x^{2} + x - 42 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} + x - 42$ 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 $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

$(x - 6) (x + 7) = 0$

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

$x - 6 = 0$

$x + 7 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 6) (x + 7) = 0$ true.

$x = 6, - 7$

Step 3

Find the Roots (Zeros) f(x)=x^2+x-42

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