Supply the numbers to write x^2+10x+16=0 in its factored form. Enter the lesser number first.(1 point) (x+() )(x+() )=0

To factor the quadratic equation \(x^2 + 10x + 16 = 0\), we need to find two numbers that multiply to \(16\) (the constant term) and add up to \(10\) (the coefficient of the linear term).

The numbers that satisfy these conditions are \(2\) and \(8\), since:

- \(2 \times 8 = 16\)
- \(2 + 8 = 10\)

Thus, the factored form of the equation is:

\[
(x + 2)(x + 8) = 0
\]

So, the answer is:

\[
(x + (2))(x + (8)) = 0
\]

If you need to enter the numbers in the format requested, you would write:

\((x + (2))(x + (8)) = 0\)

Supply the number so that the quadratic equation below has 6 and 8 as its solutions.(1 point) x^2−14x+() =0

If the quadratic equation has \(6\) and \(8\) as its solutions, we can use the fact that for a quadratic equation in the form \(x^2 - (sum\ of\ roots)x + (product\ of\ roots) = 0\), the sum and product of the roots are related to its coefficients.

1. **Sum of the Roots**:
\[
6 + 8 = 14
\]

2. **Product of the Roots**:
\[
6 \times 8 = 48
\]

Given the equation \(x^2 - 14x + () = 0\), we need to supply the number that represents the product of the roots.

Hence, you would fill in the blank with \(48\).

So the final equation would be:

\[
x^2 - 14x + 48 = 0
\]

Solve the following quadratic equation using the perfect square trinomial pattern: x^2+10x+25=0 .(1 point) x=()

To solve the quadratic equation \(x^2 + 10x + 25 = 0\) using the perfect square trinomial pattern, we can recognize that it can be factored as:

\[
(x + 5)^2 = 0
\]

This is because:

\((x + 5)(x + 5) = x^2 + 10x + 25\).

Now, solving for \(x\):

\[
(x + 5)^2 = 0
\]

Taking the square root of both sides gives us:

\[
x + 5 = 0
\]

Now, solving for \(x\):

\[
x = -5
\]

So, the solution is:

\[
x = (-5)
\]

Solve the following quadratic equation using the perfect square trinomial pattern: x^2+54x=−729 .(1 point) x=()

To solve the quadratic equation \(x^2 + 54x = -729\) using the perfect square trinomial pattern, we first move all terms to one side of the equation:

\[
x^2 + 54x + 729 = 0
\]

Next, we can identify the perfect square trinomial on the left side. The expression \(x^2 + 54x + 729\) can be rewritten as:

\[
(x + 27)^2
\]

This is because:

\[
(x + 27)(x + 27) = x^2 + 54x + 729
\]

Now we can rewrite the equation as:

\[
(x + 27)^2 = 0
\]

Next, take the square root of both sides:

\[
x + 27 = 0
\]

Now, solving for \(x\):

\[
x = -27
\]

So, the solution is:

\[
x = (-27)
\]

What are the solutions to x^2−121=0 ? Enter the lesser number first.(1 point) x=() , x=()

To solve the equation \(x^2 - 121 = 0\), we can first rewrite it as:

\[
x^2 = 121
\]

Next, we take the square root of both sides:

\[
x = \pm \sqrt{121}
\]

Since \(\sqrt{121} = 11\), we have:

\[
x = 11 \quad \text{and} \quad x = -11
\]

Now, entering the lesser number first, the solutions are:

\[
x = (-11), \, x = (11)
\]