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.
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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 think of two numbers whose sum is 10 and product is 16?
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.