These are basic factoring equations, I have the answers to all of these problems but I just need the work. I'm studying for my assessment test for college and it's very sad to say that I have forgotten how to do these basic problems. Please help. A quick explanation would be helpful, if possible, of how you got the answer.
Studying from this: http://www.crc.losrios.edu/Documents...statistics.pdf
in general, look at the problem like this: a quadratic equation of theo x^2 - Sx + P means the followings:
it has two roots x1 and x2, their sum is S and their product is P.
P = x1*x2 > if both numbers have the same sign, either both are positive, or both are negative
P < 0 if one of the numbers is positive, the other is negative. That is, they have opposite signs
S = x1 + x2 > 0 means both are positive, or the largest is positive, the other could be negative
S < 0 meas either both are negative, or the larges is negative, the other could be positive.
Once you find the roots you could write the given equation as (x - x1)(x - x2) = 0 this is what is called factoring.
Apply this to the first equation and let's see how we go from there.
Can you not see that you can factor out a "10"? And that leaves the "difference between two squares.2.
A little harder. (ax+ b)(cx+ d)= (ac)x^2+ (ad+ bc)x+ bd. Look for numbers, a, b, c, d such that ac= 9, bd= 8, and ad+ bc= 18. Try small integers. 9 factors as 3(3) while 8 factors as 2(4) so you know what numbers to try.3.
Oh, dear! Even the strongest quail before a cubic! Try some very basic facts: a cubic factors either into three linear terms: a(x- b)(x- c)(x- d) or one linear and one "irreducible" quadratic: a(x- b)(x^2+ cx+ d). In either case there is a linear factor. If (x- b) is a factor then . The "rational root theorem"- if the rational number a/b is a root then the denominator, b, must divide the leading coefficient (here, 2) and the numerator, a, must divide the constant term (here, 30). The divisors of 2 are 1, -1, 2, and -2. The divisors or 30 are 1, -1, 2, -2, 3, -3, 5, -5, 6, -6, 10, -10, 15, -15, 30, and -30. Try various "a/b" in the polynomial until you find one. (It is not necessarily true that an arbitrary polynomial has a rational zero but if it factors with integer coefficents it must have.)4.
The rest use the same ideas as above.