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Help pls (1 Viewer)
- Thread starter =)(=
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For this you sub it x=-1 into P(x) knowing the remainder is 4. You do likewise for x=1 where P(x) is now -2. You now have two simultaneous equations that are easy to solve in order to find 'a' and 'b'. That's essentially what the question is asking.
GoldenMelon
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I think this one is D, but I’m not too sure. What I did was 10C6 which is the total amount of ways to select 6 from 10 students with no restrictions and then subtract 8C4 from that which is the amount of ways you can select the teams but including those two particular students.
For this you do Total - complementary (bad case which is choosing those two particular students). Therefore the answer is 10C6 - 8C4 (since you have already chosen those two individuals).
i got a=-3 and b=1 but what do i do from there
ohh okay i thought you were allowed to have only one on the team since it said both
i did long division and got -2x +4
Lol never ever do long division for these questions. They are meant to be solved using remainder theorem. Since you have found that a=-3 and b=1 then the answer is D since dividing P(x) by (x^2-1) is the same is dividing by (x+1)(x-1) which you can see are the ones in the question. Meaning P(x) = (x+1)(x-1)Q(x) +ax+b where a = -3 and b=1 (if you're correct)
ohhh okay so the (x+1)(x-1)Q(x) becomes 0 and your left with -3x +1 ?
@=)(=, the division algorithm says that when I divide a polynomial )
by a divisor, )
, I get a quotient, )
, and a remainder, related by:
 = d(x)Q(x) + R(x))
The polynomial is divisible by if and only if  = 0)
.
In this question, we need in the form where  = x^2 - 1)
.
You can approach this by starting to look for the values
and 
given in the formula  = x^3 + 3x^2 + ax + b)
, but you actually don't need to find and unless they help you to get the polynomial in the appropriate form for the division algorithm. What you actually need to do is consider your and how to get the required form.
Now, since is of degree 2, )
can be of degree no more than 1, so let  = Ax + B)
, so you have
 = Q(x)\left(x^2 - 1\right) + Ax + B)
Note that these constants
and 
are not the same as and , so perhaps different symbols might be wise.
In any case, we are given that the remainder is -4 on division by)
, so  = 4)
by the remainder theorem:
 &= 4 \\ Q(-1)\left[(-1)^2 - 1\right] + A(-1) + B &= 4 \\ Q(-1) \times 0 - A + B &= 4 \\ B - A &= 4 \end{align*})
Also, on division by)
, the remainder is -2, so  = -2)
by the remainder theorem.
 &= -2 \\ Q(1)\left[(1)^2 - 1\right] + A(1) + B &= -2 \\ Q(1) \times 0 + A + B &= -2 \\ A + B &= -2 \end{align*})
Solving simultaneously, we get
and 
, and so  = \left(x^2 - 1\right)Q(x) - 3x + 1)
and so the remainder on division by )
is 
and the answer is D.
The polynomial is divisible by
In this question, we need
You can approach this by starting to look for the values
Now, since
Note that these constants
In any case, we are given that the remainder is -4 on division by
Also, on division by
Solving simultaneously, we get
Just like if I divide 73 by 10, I get 7 with a remainder of 3, that is
Starting number = divisor x quotient + remainder