Not sure about remainder theorem but I'm sure that the last terms should all multiply tho the last term
see the expanded form is -45
so the last terms of each binomial should multiply to -45
3 times -5 times ?=-45
-15 times ?=-45
divide by -15
?=3
the question mark is 3
Answer:
40
Step-by-step explanation:
(2x+1/(2x))^5 *(2x -1/(2x))^5
= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2
= (4x^2-1/4(x)^2)^5
now
x =4x^2. ,a = 1/4(x)^2 ,n =5
we have
general term = Cr *x^r *a^(n-r)
= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)
= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)
= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)
= Cr * 4(2r-n) *x(4r-2n)
now for x^2
4r-2n = 2
4r -10=2
4r =12
r = 3
now for coeff
C(5,3) * 4^(2*3-5)
5!/(3!*(5-3)!) * 4
5*4/(2*1)*4
40
That would be 8 and 9.
9*8=72
While 9+8=17
I hope this helps! (:
<span>Multiply the number in the tens place of the bottom number by the number in hundreds place of the top number. Multiply 3 times 7 to equal 21. Add the 1 you carried to equal 22. You don't need to carry the 2 in 22, as there are no more numbers to multiply on this line, so you can just write it down next to the 6.</span>
