If guessing does not work, "completing the square" will do the job.

Move the constant term to the other side of the equation:

The magic trick of this method is to exploit the binomial formula:

If we look at the left side of the equation we want to solve, we see that it matches the first two terms of the binomial formula if *b*=1. Let's write down the binomial formula for *b*=1:

But the third term of the binomial formula does not show up in our equation; we make it show up by force by adding 1 to both sides of our equation:

This trick is called "completing the square"! Now we use the binomial formula to simplify the left side of our equation (also adding 7+1=8):

Next we take square roots of both sides, but be careful: there are **two** possible cases:

In both cases .
We are done, once we solve the two equations for *x*.

are the two roots of our polynomial. Consequently, our polynomial factors as follows:

Let us try to factor . We will again consider the equivalent problem of finding the roots, the solutions of the equation

In this example, the leading coefficient (the number in front of the ) is 2, and thus not equal to 1; we can fix that by dividing by 2:

Move the constant term to the other side of the equation:

The magic trick of this method is to exploit the binomial formula:

If we look at the left side of the equation we want to solve, we see that it matches the first two terms of the binomial formula if *b*=3/2. Let's write down the binomial formula for *b*=3/2:

But the third term of the binomial formula does not show up in our equation; we make it show up by adding to both sides of our equation:

We have "completed the square"! Now we use the binomial formula to simplify the left side of our equation (also simplifying the right side):

The rest is easy: we take square roots of both sides, but be careful: there are **two** possible cases:

In both cases .
We are done, once we solve the two equations for *x*.

are the two roots of our polynomial. Here is the factorization of our polynomial. Be careful: We have to multiply our two "standard" factors by 2 (the leading term we divided by in the beginning!)

Let us factor . We will again consider the equivalent problem of finding the roots, the solutions of the equation

Move the constant term to the other side of the equation:

The magic trick of this method is to exploit the binomial formula:

If we look at the left side of the equation we want to solve, we see that it matches the first two terms of the binomial formula if *b*=-3. **You always take half of the term in front of the x. **Let's write down the binomial formula for

But the third term of the binomial formula does not show up in our equation; we make it show up by adding 9 to both sides of our equation:

We have "completed the square"! Now we use the binomial formula to simplify the left side of our equation (also simplifying the right side):

The rest is easy: we take square roots of both sides, but be careful: there are **two** possible cases:

In both cases .
We are done, once we solve the two equations for *x*.

are the two roots of our polynomial. Here is the factorization of our polynomial.

There are more exercises on the next page dealing with the quadratic formula.

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