Hi I'm Margarette Xiel Ehne Medrano and i'm blogging about what we learned on today's class. We learned about Factoring to find zero. In Factoring to find zero, if a quadratic is factorable, it is easy to find the zeros(x-intercepts) of the function.

Example: Find the x-intercept of f(x)= x^2+8x+15

(x+5)(x+3)=f(x)

(x+5)(x+3)=0

x+5=0 x+3=0

x=-5 x=-3

When a quadtratic is not factorable, sometimes it is possible to find the zeros by completing the square and then solving for x. It is the best to use this method under the following condition:

1)The quadtratic does not factor

2)The middle term is even

3)The value of k is negative(you will have to complete the square first to know this information)

note: if not factorable just complete the squares

Example:Find the zeros of y=(x+3)^2-14

0=(x+3)^2-14

*transpose and square root both sides

14=(x+3)^2

14=x+3

x+3=+-14

This lesson covers many ways to solve quadratics, such as taking square roots, completing the square, and using the Quadratic Formula. But we'll start with solving by factoring.

You should already know how to factor quadratics. (If not, review Factoring Quadratics.) The new thing here is that the quadratic is part of an equation, and you're told to solve for the values of x that make the equation true. Here's how it works:

• Solve (x – 3)(x – 4) = 0.

Okay, this one is already factored for me. But how do I solve this?

Think: If I multiply two things together and the result is zero, what can I say about those two things? I can say that at least one of them must also be zero. That is, the only way to multiply and get zero is to multiply by zero. (This is sometimes called "The Zero Factor Property" or "Rule" or "Principle".)

Warning: You cannot make this statement about any other number! You can only make the conclusion about the factors ("one of them must equal zero") if the product itself equals zero. If the above product of factors had been equal to, say, 4, then we would still have no idea what was the value of either of the factors; we would not have been able (we would not have been mathematically "justified") in making anyclaim about the values of the factors. Because you can only make the conclusion ("one of the factors must have equalled zero") if the product equals zero, you must always have the equation in the form "(quadratic) equals (zero)" before you can attempt to solve it.

The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, I'll set them each equal to zero:

x – 3 = 0 or x – 4 = 0

This gives me simple linear equations, and they're easy to solve:

x = 3 or x = 4

And this is the solution they're looking for: x = 3, 4

Note that "x = 3, 4" means the same thing as "x = 3 or x = 4"; the only difference is the formatting. The "x = 3, 4" format is more-typically used.

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One important issue should be mentioned at this point: Just as with linear equations, the solutions to quadratic equations may be verified by plugging them back into the original equation, and making sure that they work, that they result in a true statement. For the above example, we would do the following:

Checking x = 3 in (x – 3)(x – 4) = 0:

([3] – 3)([3] – 4) ?=? 0
(3 – 3)(3 – 4) ?=? 0
(0)(–1) ?=? 0
0 = 0

Checking x = 4 in (x – 3)(x – 4) = 0:

([4] – 3)([4] – 4) ?=? 0
(4 – 3)(4 – 4) ?=? 0
(1)(0) ?=? 0
0 = 0

So both solutions "check" and are thus verified as being correct.

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• Solve x² + 5x + 6 = 0.

This equation is already in the form "(quadratic) equals (zero)" but, unlike the previous example, this isn't yet factored. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.

So the first thing I have to do is factor:

x² + 5x + 6 = (x + 2)(x + 3)

Set this equal to zero:

(x + 2)(x + 3) = 0

Solve each factor: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

x + 2 = 0 or x + 3 = 0
x = –2 or x = – 3

The solution to x² + 5x + 6 = 0 is x = –3, –2

Checking x = –3 and x = –2 in x² + 5x + 6 = 0:

[–3]² + 5[–3] + 6 ?=? 0
9 – 15 + 6 ?=? 0
9 + 6 – 15 ?=? 0
15 – 15 ?=? 0
0 = 0

[–2]² + 5[–2] + 6 ?=? 0
4 – 10 + 6 ?=? 0
4 + 6 – 10 ?=? 0
10 – 10 ?=? 0
0 = 0

So both solutions "check".

I hope this summary of today's class helped you(: