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# More examples of factoring quadratics as (x+a)(x+b)

Can't get enough of Sal factoring simple quadratics? Here's a handful of examples just for you! Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

• Can both a and b equal the same for example x^2+20x+100 can both a and b equal 10?
• Yes, a and b can be equal. In the example you wrote, that expression is a special case. It is the square of a binomial. (x+10)*(x+10) = (x+10)^2. x+10 is a binomial that is being squared. I hope I helped.
• What's a coeffcient?
• it is the number that is multiplied by a variable
• how come the constant term is the product of a and b?
• Let me try (and please know that Sal already tried at )...
Because when I you have a quadratic in intercept form `(x+a)(x+b)` like so, and you factor it (basically meaning multiply it and undo it into slandered form) you get: `x^2 + bx + ax + ab`. This of course can be combined to: `x^2 + (a+b)x + ab`. So when you write out a problem like the one he had at `x^2 + 15x + 50`, 50, which is your "C" term (the third term) and is your constant, 50 is the product of a and b (ab). This can be shown here:
`` x^2 + 15x + 50 is equal to:(x+5)(x+10) =x^2 + 10x + 5x + 50 =x^2 + 15x + 50 {which is what we started with.}``

Thank you very much for asking this question, i was wondering it for a long time and now that I know it I am glad that I am not the only one who was confused and I am glad to be of possible service. Please let me know if this helped anyone.
• What if the numbers are prime and have no factors?
• All quadratics can be factored, but not all of them can be factored with rational numbers or even real numbers. If a quadratic cannot be factored into rational factors, it is said to be irreducible. However, it is always possible to factor a quadratic, if you allow irrational or complex factors.

Here's how to factor ANY quadratic expression in the form: ax² + bx+c.
Let d = b² - 4ac
(If d is not a positive perfect square, then the quadratic is "irreducible".)
The factors are:
a [x + ¹⁄₂ₐ (b + √d)] [x + ¹⁄₂ₐ (b - √d)]
If it makes for convenient numbers, you may use the distributive property to multiply a into one (but not both) of the factors.
• I am confused. Is the coefficient the number that the factors are added/subtracted by?
• No because the coefficient shows how many you have of a variable
• at we do not have to multiply negative 1 to both (x-3)(x+8) ?
• No, because -x*-x would equal positive x².
The goal is to get -x².
So, -x*x = -x²
Hope this helps!
• At , can you do (-x-3)(x+8) instead of -1(x-3)(x+8)?
• No. The distributive property is not being properly used.

You are trying to multiply -1 by (x-3). Using the distributive property, you'll get -x+3 as a result. So it should be (-x+3)(x+8).
• I was wondering... I know a Quadratic Expression is a second degree expression... but doesn't "quad" mean 4?? Also are there other names for other expression, like for e.g. third degree or fourth degree expressions? Thanks if you answered! `:D`
• Because quadratus is the Latin for "square" due to there being four sides on a square.

The second power of a number is called its square because if we have an integer, and construct a square with that number of items on each side, the total number will be its second power. E.g. a 4×4 square having 16 items:

* * * *
* * * *
* * * *
* * * *
I found this explanation on Google: