um math prep: s7.3 - solving by factoring

Some equations can be put into factored form such that the product of the factors is zero. Then we can solve by using the fact that if (a)(b) = 0, then either a or b (or both) is zero.

Examples
For these examples, notice that we first get the equation into the form (a)(b) = 0 and then find the zeros. See the cautionary notes, below, discussing this.

Solve for x:
Solve for x:
Solve:

is never zero, so x=-1 is the only solution.

Be careful not to make these common mistakes:

When solving

do not make the mistake of setting x + 1 =15 and x + 3 = 15. It is not true that (a)(b) = 15 means that a = 15 or b = 15 (or both). Thus we have to expand the left-hand side and set the equation to zero.
When solving
,
we might be tempted to divide both sides by x + 3. However, if we do this we omit the solution x = -3, because we can't divide by zero.

Solving with the quadratic formula

If we are unable to factor an equation and it is quadratic (the highest power of the variable is 2) we can use the quadratic formula. If

then the solutions are

Examples

Solve for x:

This is the same as

,
which we are unable to factor using integers, so we use
x=(-2 +/- sqrt(4 - 4(-1)(11)))/(2(-1)) = (-2 +/- sqrt(48))/(-2) = 1 +/- 2sqrt(3)

Note that we can approximate these with the decimal approximations

and

. We could find these directly from a graph or a calculator.

Practice

Be sure you've gone through each step in the examples above before doing these. Once you've worked them until you're sure that you understand them, go on to the next section. There are more problems of this type in the section test at the end of the section.

Note that you can get new practice problems by clicking the "Refresh" button at the bottom of the practice set.

Solve for x:

Answer: and

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Solve for x:

Answer: and

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Solving by Factoring

Solving with the quadratic formula