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Factoring quadratics as (x+a)(x+b) (example 2)

Sal factors x^2-14x+40 as (x-4)(x-10) and x^2-x-12 as (x+3)(x-4). Created by Sal Khan.

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• I didn't get
• The product of two numbers just means what you get when the two numbers are multiplied together. He's looking for two numbers a and b such that a + b (their sum) is -14 and ab (their product) is 40.
• Sal says that this is guess work, but:
a+b = -14 and ab = 40
looks like a very obvious 'systems of equations' problem. When I tried it though things got messy very quickly. Neither substitution nor elimination worked. Can anyone explain why?
• This arises from a quadratic, and if you try to solve for either a or b, you will get a quadratic again. The best procedure is just to test factors of 40 until you realize that -4 and -10 are legitimate solutions to this system.
• why do they call it a quadratic?
• Since the highest exponent is a 2. `x^2 + 2x - 3`.
• how can I deeply learn this some more. also where in my life would I have to use this besides math class lol!! some examples might help
• yo tobias i ask the same question everyday, yet everyday i tell myself i need this credit!
• What is the purpose in Factoring?
• It can help us find roots of polynomials, and it can help us simplify rational expressions.
• how would you factor a value for x^2 where the coefficient is greater than 1?
• Simply put, we would just change the value that we need a and b to multiply to. Currently, we always try to get a and b to multiply to whatever the last term in our quadratic is. But really what we're looking for is for our a and b to multiply to the coefficient of the first term and the value of the last term. Here's a better example.

All quadratics are written in the form:

ax^2 + bx + c

(In this case, a and b have no relation to the a and b that Sal is talking about for factoring.)

Based on this equation, we want our two factors to multiply to a*c.

For example, if we have the equation:

4x^2 + 9x + 10

Then we want our a and b to multiply to 4*10.

Hope that helps a little!
• I apologize for not being very specific, but does anyone know a faster way to factor? This is the same way we learned it in school and I would like to know some formulas, etc., to factor polynomials like these. Thanks!!
• I don't think there are any formulas to factor polynomials, but there always is the quadratic formula. This is any easy way of finding roots (x-intercepts) of a quadratic equation by just plugging values into an equation.
• does the order of the (x )(x ) matter if they are the same coefficients with the correct sign? say the solution is (x-2)(x+3). would it matter if it was (x+3)(x-2)?
• The order does not matter because the commutative property of multiplication applies. This is just like 2(3) is the same as 3(2).
Hope this helps.
• What if you have something that can't be factored?
• Then you leave it as it is. For exmple `x^2 + 6x + 23` can not be factored because no two numbers multiplied together will equal 23.
• (3x + 4y)^2 + 2(3x +4y)(3x-4y) + (3x-4y)^2

Can this be factorised?

How?
• There are two ways to handle this. You can substitute `v = 3x+4y` and `w = 3x - 4y`, then you have
`v² + 2 v w + w² = (v + w)(v + w) = (v + w)²`
`(v + w)² = (3x + 4y + 3x - 4y)² = (6x)² = 36x²`
`(3x + 4y)² + 2(3x + 4y)(3x - 4y) + (3x - 4y)² = (9x² + 24xy + 16y²) + (18x² - 32y²) + (9x² - 24xy + 16y²)= 9x² + 18x² + 9x² + 24xy - 24xy + 16y² -32y² + 16 y² = 36x²`