Below are two ways of doing it. The first one given is a "brute force" method, while the second is neater.
Notation: a, b, c, p, q, r and s are constants dependent on the question, while x is a variable. My 2 is your ^2, and when I write x1 or x2, these are the solutions to the "quadratic formula" when you either add the 'square root bit' or take it away.If you can't immediately factor polynomials (these are actually binomials which makes factorising easier), the fastest way to factorise is to solve the equation ax
2+bx+c=0 and then use the quadratic formula:
Now wrote the polynomial as (x-x
1)(x-x
2) where x
1 and x
2 are the two values of x you got in the formula above.
For the specific problems you have given:
1) d is your variable x, a is 36, b is 5 and c is -24. You could then plug those in to the formula above.
2) treating (a-b) as x, a is 10, b is -11 and c is -6.
3) Now the first thing to do here is work out 14(2-x)
2 - 15(2-x)-11 as a binomial in x rather than (2-x). Since 14(2-x)
2= 14(4-4x+x
2) and -15(2-x) = 15x-30 you can express this polynomial as 14x
2-41x+15, then solve as before.
The above method works but is very tedious and not particularly insightful. Let's approach the problem in a new way.
Let's suppose that you have already factorised the equation ax
2+bx+c. You'll get two brackets, each linear in x, of the form (px+q)(rx+s). Now what we do is expand the brackets again! We find that (px+q)(rx+s) = (pr)x
2+(ps+qr)x+(qs). Now we already know, though, that:
(px+q)(rx+s)=ax
2+bx+c
There's an important theorem that says that two polynomials are the same if they have the same coefficients (which are the numbers a, b and c in ax
2+bx+c). So we conclude that:
pr=a
qr+ps=b
qs=c
This amounts, annoyingly, to three equations with four unknowns. There's not much, then, that can be done to solve these equations but if these were homework questions it's likely that they will factorise nicely. So now we start trialing whole numbers that divide a and c and compare them with b to see if we can get a match.
For example, look at question one. We know that there are two factors (pd+q)(rd+s) to make the equation 36d
2-5d-24. Hence:
pr=36
qr+ps= -5
qs=-24.
Let's try q=3, s= -8, r=9, p=4. By some strange coincidence (
) we find that 9*4 is 36, 3*-8 is -24, and 3+9+4*-8 = -5. So this works! Hence the solution to the first question is 36d
2-5d-24=(4d+3)(9d-8).
Try the same sort of things for questions 2 and 3, always trying to pick whole numbers that divide into the numbers given. So in the 3rd question you're looking for two numbers that divide -11 and two that divide 14. If you can't find any (which this time is a possibility since questions in homework are like that) then against expand (2-x)
2 (It's easiest to take (2-x) as the variable if you are doing things algebraically, then note that if, say, one bracket has the form
(p(2-x)-q), then you can write this as (-px+2p-q).)
If this isn't clear to you then I'll try to explain further later.
