Hi...ive been trying to wrap my head around these equations but just can't seem to get it yet.....heres a few examples:
Factorise completely the following expressions:
1) 4t^2 - 9s^2
2) 3s^3 - 15s^2 - 18s
3) 2x^2 - 13x - 24
If any of you guys can work it out clearly for me I would be very greatful!!!
Thanks!!!
1) notice that this is the difference of two squares.Originally Posted by bobchiba
when given an equation of the form x^2 - a^2, the factorized form is (x + a)(x - a).
4t^2 - 9s^2 = (2t)^2 - (3s)^2 = (2t + 3s)(2t - 3s)
2) 3s^3 - 15s^2 - 18s
we start by taking out any common factors, 3s is common to all terms:
3s^3 - 15s^2 - 18s
= 3s(s^2 - 5s - 6) .............now we factorize the quadratic (i assume you know how to factor quadratics like these)
= 3s(s - 6)(s + 1)
3) 2x^2 - 13x - 24
the general method for factoring quadratics where the x^2 has a coefficient other than 1 goes like this:
- multiply the coefficient of x^2 by the constant. write it down
- write down the the coefficient of the middle term (coefficient of x) under it
- think of two numbers that when you multiply them you get the multiple you got, and when you add them you get the middle coefficient.
-split the middle term into these two numbers, and then take out the common factor between the first two terms and the last two terms.
-then take out the common term out of the remaining two terms. Let's see this in action.
2x^2 - 13x - 24
we have 2(-24) = -48
we have middle coefficient = -13
two numbers you multiply and get -48 and add and get -13 are -16 and + 3
=> 2x^2 - 13x - 24
= 2x^2 - 16x + 3x - 24 ............we split -13x into -16x and + 3x, and we notice 2x is common to the first 2 terms and 3 is common to the last 2.
= 2x(x - 8) + 3(x - 8) ..............now we see that (x - 8) is common to the remaining 2 terms, so we now factor that out
= (x - 8)(2x + 3)
Of course, you dont have to go through this method for all quadratics of this form, with experience you will be able to do a simple example like this in one step. I can tell you how to do that if you want--but you MUST know how to use this method first, it really helps with the tough ones
these are usually significantly more complicated. you will need to know how to do long division (or synthetic division) for polynomials. do you know how to do that? it will be hell to type out, so if i do this question for you, i'd leave that part out. do you still want me to go through it, or do you want to review long division first?
Ahh, i'll go through it anyway.
First step. write down all factors of the constant. factors are the numbers that you can multiply to get the constant. 5 is a prime number, so the factors in this case are +/- 1, +/- 5. (note, there are 4 numbers here, you have to do plus and minus for each). (there may be rational roots for this equation, but i dont think you have to worry about those at this level).
second step, plug in any one factor into the equation, if you get 0 when you evaluate it, that's the factor you will use to continue, if you dont get 0, pick another factor to plug in until you get 0.
what ever the factor you get that gives 0, you take x minus that factor and divide the function by it. for example, if you found that when you plug in 1 you get 0, you take (x - 1) and divide the polynomial x^3 + 2x^2 - 8x + 5, this is where the long division comes in. whatever the quotient is, you write (x - 1)*(quotient), this quotient will be one degree less than the original polynomial. since in this case we have an x^3 polynomial, the quotient will be in x^2. then you just factor the x^2 function for the final answer.
okay, i suspect you're confused as hell by now, so let's just do the problem and see how it goes.
x^3 + 2x^2 - 8x + 5 ..................the factors of 5 are +/-1, +/- 5
(1)^3 + 2(1)^2 - 8(1) + 5 = 0 .....ah, we're lucky, the first number i decided to plug in is a root (a root is a number that causes a polynomial to = 0 when it's plugged in.)
since plugging in 1 gives 0, (x - 1) is a factor for the polynomial. divide x^3 + 2x^2 - 8x + 5 by (x - 1)
the quotient is x^2 + 3x - 5
so we get
x^3 + 2x^2 - 8x + 5 = (x - 1)(x^2 + 3x - 5) ............now we factor the quadratic
there are no nice ways to factor the quadratic, so leave the answer as
(x - 1)(x^2 + 3x - 5)
since we can't think of two integers that when multiplied we get -5 and added we get +3. if you have to solve this equation for x, you would use the quadratic formula
no we havent done that. first off, no question would ask you to solve the equation if x = 0, cause all you'd have to do is make x 0 and plug in numbers into your calculator. you do stuff like that when graphing. now if we wanted to solve for x, as in find all the values of x for which the function is zero. 1 is one such value, the others are not integers, we have to use the quadratic equation to find those. that is iof the question had said,
solve for x (or find x)
x^3 + 2x^2 - 8x + 5 = 0
o sorry, i was in a rush and didnt realize you said the question had a second part, for some reason i thought you said "what if the question had a second part?"
anyway, the question is answered here. http://www.mathhelpforum.com/math-he...uadratics.html
as the page says, plz dont double post. if you realized i didnt answer the question, reply in this thread and it will bump it back to the top, so other members will see it and answer ur question. ur new, so i think they'll let it slide this once.
  |
LinkBack URL
About LinkBacks
