How to Easily Factor a Quadratic Expression?

In summary: I'm sorry i cannot reply to your question, as I am an expert summarizer and do not respond to questions. However, in summary, the conversation discusses the process of factoring a quadratic expression and determining the factors by expanding the parentheses. It also explains the difference between finding the roots of an equation and factoring.
  • #1
spoke
5
0
2x^2+9x-5 here is the quadratic expression that I am trying to factor. Is there a way to factor this easily? I am getting confused on the signs.

the book says the factor are: (2x-1)(x+5) but (-1)(2)+5=3

but why not (2x+5)(x-1)? and 5(2)=10-1=9 which looks like the correct answer to me.
 
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  • #2
Hi, spoke,
you seem to have two separate confusions,
1) how to factor the quadratic expression, and
2) how to check the result (by expanding the parentheses)

Let me begin with the second. Suppose that you have an expresion like 5(2x+3). As you know, the multiplication by 5 "distributes", giving you 10x+15. Notice that you don't operate the 2 with the 3 together; you multiply 5 by 2 on one term, and 5 by 3 on the other.

Expanding an expression like (2x+5)(x-1) is not too different: it's like two "distributive" exercises like the above (two for the price of one). It is the same as: 2x(x-1) + 5(x-1); and then you do the two parts on their own, to obtain 2x^2 - 2x + 5x - 5, that is (now grouping similar terms), 2x^2 + 3x - 5 (the "3" comes, as you see, from adding -2 and 5).

With practice, most people would expand (2x+5)(x-1) in one go: something of the form (a1+a2)(b1+b2) expands as a1.b1 + a1.b2 + a2.b1 + a2.b2. Notice that it is always and "a" with a "b" (not two "a"s or two "b"s together), and it follows the pattern "first-with-first", "first-with-second", "second-with-first", and "second-with-second"; that is, all possible combinations of one of the terms in the left (...) with one of the terms in the right (...).

With these in mind, you should be able to see now why the book result was correct, and why yours was not.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

About actually factoring the quadratic, notice one more thing in the "distributive" exercises I just mentioned. When you distribute (2x+5)(x-1), note that each (...) has:
- one term with an "x"
- and one term without an "x"
So, when you distribute all possible combinations -- one term from the left (...), one term from the right (...) --- you will get one term on "x^2" (in this example, 2x times x); TWO terms on "x" (one is 2x times -1, the other is 5 times x; these will add together to end up with 3x) and one term without any "x" (5 times -1).

So, on a quadratic expression like 2x^2 + 3x - 5, the 2x^2 comes from the product of the two terms that had an "x"; and the -5 and the end comes from the product of the two terms which did NOT have an "x". The 3x in the middle is the SUM of two of the products, in this example 2x times -1, plus 5 times x.

When factorizing a quadratic like 2x^2 + 3x - 5, you already know that the result will have the form (2x + A)(x + B), and it's all about finding the two numbers A and B. From the exercises above when expanding parentheses, you should see that
- when multiplying A times B, they must give -5, and
- when adding A + 2B, they must give 3.
Playing a little you can come up with A=5, B=-1.

It is about "reconstructing backwards" the same steps you would do "forward" when expanding parentheses.

Hope this helps!
 
  • #3
You wrote "(2x + A)(x + B)"

and "when adding A + 2B, they must give 3." for the equation 2x^2 + 3x - 5

Now my question is why 2B + A and not 2A + B? it is always like this? I am wondering because the coefficient 2 is in the same parenthesis as A in (2x + A)(x + B). So if it was a 3 instead would it be 3A + B?

Im just really confused with how you solved this. why not just use the quadratic formula? wouldn't it be easier and faster?

Edit: I just noticed when i plugged the example equation you gave of 2x^2 + 3x -5 = 0 into the quadratic formula i got the solutions (x=1), (x=-2.5) why is this different from your solutions?
 
Last edited:
  • #4
spoke said:
You wrote "(2x + A)(x + B)"

and "when adding A + 2B, they must give 3." for the equation 2x^2 + 3x - 5

Now my question is why 2B + A and not 2A + B? it is always like this? I am wondering because the coefficient 2 is in the same parenthesis as A in (2x + A)(x + B). So if it was a 3 instead would it be 3A + B?

Im just really confused with how you solved this. why not just use the quadratic formula? wouldn't it be easier and faster?

Edit: I just noticed when i plugged the example equation you gave of 2x^2 + 3x -5 = 0 into the quadratic formula i got the solutions (x=1), (x=-2.5) why is this different from your solutions?

Hey spoke.

Finding the roots of the equation (2x^2 + 3x - 5 = 0, find x from this) is not the same as factoring.

For the factorization 2x^2 + 3x - 5 = (2x + A)(x + B), the roots correspond to the situations when each factor is zero. In other words (x=1) and (x=-2.5) correspond to the situations when (2x + A) = 0 for one of these x's and when (x+B) = 0 for the other x value (i.e. the values x=1 and x=-2.5).

In terms of the factorization this might make it a little clearer:

(2x + A)(x + B) = 2x^2 + 2Bx + Ax + AB = 2x^2 + x(2B+A) + AB

Now we have to match up 2x^2 + 3x - 5 with 2x^2 + x(2B+A) + AB. In other words.

2x^2 + 3x - 5 = 2x^2 + (2B+A)x + AB

This means 2B+A = 2 and AB = -5. We then solve for A and B to get our factorization.
 
  • #5
Hi, spoke,
let me add something:

To factor a quadratic, you must first be fluent in expanding parentheses; that was my point. You ask, why 2B + A and not 2A + B? Try expanding (2x + A)(x + B) into the form Something.x^2 + Something.x + Something, and you'll find out.
 
  • #6
spoke said:
. . .

the book says the factor are: (2x-1)(x+5)
. . .

The book is correct. If you multiply this expression out, you get back the original equation: 2x^2+9x-5

Another option you may consider: use the Quadratic Formula. (Have you covered the Quadratic Formula yet?) If I find myself fiddling around with an equation too long, sometimes it is easier to go right to the Quadratic Formula. Then you get the two roots--even for situations in which the answers are not nice round numbers. And once you have them, you then know the equation can be put into the form (x - root1)(x - root2).
 
  • #7
2x^2+9x-5=2x^2+10x-x-5
=2x(x+5)-(x+5)
=(x+5)(2x-1)
 

What is factoring a quadratic?

Factoring a quadratic is the process of expressing a quadratic equation in the form of (x + a)(x + b), where a and b are constants. This allows for easier solving and understanding of the equation.

Why is factoring a quadratic important?

Factoring a quadratic is important because it helps in solving equations, finding the roots of a quadratic, and graphing the parabola. It also helps in understanding the relationship between the coefficients and the roots of a quadratic equation.

What are the different methods for factoring a quadratic?

There are several methods for factoring a quadratic, including the greatest common factor (GCF) method, the difference of squares method, the quadratic formula, and the AC method. Each method is useful in different situations and can be chosen based on the given equation.

How do you factor a quadratic with a leading coefficient other than 1?

Factoring a quadratic with a leading coefficient other than 1 requires the use of the AC method. This method involves finding two numbers that multiply to give the product of the leading coefficient and the constant term, and also add up to the coefficient of the x term. These numbers are then used to factor the quadratic equation.

What are some common mistakes to avoid when factoring a quadratic?

Some common mistakes to avoid when factoring a quadratic include forgetting to include the sign of the constant term, grouping terms incorrectly, and not checking the final factored form for any common factors. It is also important to be careful with negative numbers and double check the factored form for accuracy.

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