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I need help factoring.

  1. Jun 21, 2006 #1
    I haven't taken a math class since high school and I'm 23 now. I jumped right into Precalc 1 for the summer and the first chapter kicked my ass. I completely forgot how to factor and do LCD with algebraic equations and my professor just breezes by it like its nothing.

    Can anyone explain the general principles of factoring trinomial equations? Example x^2 + 5x 6 = 0, factor that. (Or is the term called FOIL?)

    LCD is a problem too, I know the concept is multiply both denominators together to get the LCD to cancel out... but what if the equation is complex like (2+5x)/(x^2+4x+3) + (3-4x)/(5x+8) = 7x^2 + 7

    another one was x/8 + 2x/4 = 25

    Any help and or practice problems would be greatly appreciated, i need to catch up!
     
  2. jcsd
  3. Jun 22, 2006 #2

    HallsofIvy

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    FOIL is a mnemonic for multiplying, the opposite of factoring. To multiply (x+a)(x+b) you have to multiply each part of the first term by each part of the second term. F is "first terms" x*x= x2. O is "outside" x(b)= bx, I is "inside" a(x)= ax, L is "last" a(b)= ab, but the effect is simply that you multiplied the "x" in the first term by both x and b and the "a" in the first term by both x and b: (x+a)(x+b)= x2+ ax+ bx+ ab= x2+ (a+b)x+ ab. Now look carefully at the number multiplying x, a+ b, and the constant term ab. They are just the sum and product of the two numbers. To factor x2 + 5x+ 6, work backwards. You know that ab must equal 6 so factor the 6 first: 6= 3(2). It also happens that 5= 3+2= a+b. x2+ 5x+ 6= (x+3)(x+2).
    You do not "multiply both denominators together to get the LCD to cancel out". The point of the LCD "Least common denominator" is to not have to do that much work (unless absolutely necessary. To solve
    [tex]\frac{2+ 5x}{x^2+4x+3}+\frac{3-4x}{5x+8}= 7x^2+ 7[/tex]
    factor the denominators: 3= 3(1) and 3+1= 4 so x2+ 4x+ 3= (x+3)(x+1). Since 5x+8 is not either of those, the LCD does happen to be the product: multiply each term of the equation by x2+ 4x+ 3 and 5x+ 8. Each of the denominators will cancel and you will have
    [tex](2+5x)(5x+8)+ (3-4x)(x^2+ 4x+ 3)= (7x^2+ 7)(x^2+ 4x+ 3)(5x+8)[/itex]
    The will be a fifth degree equation which might be impossible to solve exactly. I assume you just made that up, it's not an actual problem you were expected to solve.

    For x/8+ 2x/4= 25, since 8= 2(4), just multiply each term by the LCD,8:
    8(x/8)+ 8(2x)/4= 8(25), x+ 4x= 200, 5x= 200.
     
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