Register to reply 
Factoring a quadratic 
Share this thread: 
#1
Apr612, 01:35 AM

P: 5

2x^2+9x5 here is the quadratic expression that im trying to factor. Is there a way to factor this easily? im getting confused on the signs.
the book says the factor are: (2x1)(x+5) but (1)(2)+5=3 but why not (2x+5)(x1)? and 5(2)=101=9 which looks like the correct answer to me. 


#2
Apr612, 03:23 AM

P: 688

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)(x1) 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(x1) + 5(x1); 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)(x1) 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 "firstwithfirst", "firstwithsecond", "secondwithfirst", and "secondwithsecond"; 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)(x1), 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
Apr712, 12:08 AM

P: 5

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? im 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? wouldnt 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? 


#4
Apr712, 02:05 AM

P: 4,572

Factoring a quadratic
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
Apr712, 03:12 AM

P: 688

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
Apr712, 11:47 AM

P: 90

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 rootseven 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
Apr812, 12:54 PM

P: 2

2x^2+9x5=2x^2+10xx5
=2x(x+5)(x+5) =(x+5)(2x1) 


Register to reply 
Related Discussions  
Factoring question  generalized factoring in integers  Linear & Abstract Algebra  0  
Solving Quadratic Equation Using Box Factoring  Precalculus Mathematics Homework  2  
Factoring a quadratic  Precalculus Mathematics Homework  7  
Quadratic equations and inequalities / applications of quadratic functions question  Precalculus Mathematics Homework  3  
Factoring with the quadratic formula  Precalculus Mathematics Homework  2 