F. 2 Maths (最佳解答取30分)

1) -243ac^2+75b^2a
-243ac^2+75b^2a
=-81* 3a*c^2+25* 3a* b^2... find out the common factors 3a.
= 3a* (-81 * c^2 + 25b^2) ..... take out the common factors 3a.
= 3a* (25b^2-81 * c^2 ) ..... ..re-arrange the terms
= 3a* ((5b)^2-(9 c)^2 ) ..... completing square
= 3a* (5b+9c)(5b-9 c)..... difference of two squares (A^2-B^2)=(A+B)(A-B)
done!

2) p^2-(p+q)^2

= (p+p+q)(p-(p+q))..... difference of two squares (A^2-B^2)=(A+B)(A-B)
=(2p+q)(-q)
=-q(2p+q)
3) 25(x+1)^2-16(x-1)^2
=5^2(x+1)^2-4^2(x-1)^2
=(5x+5)^2-(4x-4)^2
= (5x+5+4x-4)(5x+5-(4x-4))..... difference of two squares
=(9x+1)(5x+5-4x+4)
=(9x+1)(x+9)

1) -243ac^2+75b^2a
ans:
-243ac^2+75b^2a
=-3a(81c^2-25b^2)<----take out -3a
=-3a((9c)^2-(5b)^2)<----81=9^2,25=5^2
=-3a(9c-5b)(9c+5b)<---using identity a^2-b^2=(a-b)(a+b)

2) p^2-(p+q)^2
ans:
p^2-(p+q)^2
=(p-(p+q))(p+(p+q))<---using identity a^2-b^2=(a-b)(a+b)
=-q(2p+q)

3) 25(x+1)^2-16(x-1)^2
ans:
25(x+1)^2-16(x-1)^2
=(5(x+1))^2-(4(x-1)^2)<----25=5^2,16=4^2
=(5x+5)^2-(4x-4)^2
=(5x+5-(4x-4))(5x+5+(4x-4))<---using identity a^2-b^2=(a-b)(a+b)
=(x+9)(9x+1)


Is it detail enough?
Hope I can help you^^

1) -243ac^2+75b^2a
=-3a(81c^2+25b^2)
=-3a(9c+5b)(9c-5b) ==> a^2-b^2=(a+b)(a-b)

2) p^2-(p+q)^2
=[p-(p+q)][p+(p+q)] ==> a^2-b^2=(a+b)(a-b)
=(p-p-q)(p+p+q)
=-q(2p+q)

3) 25(x+1)^2-16(x-1)^2
=[5(x+1)-4(x-1)][5(x+1)+4(x-1)] ==> a^2-b^2=(a+b)(a-b)
=(5x+5-4x+4)(5x+5+4x-4)
=(x+9)(9x+1)

(1) -243ac^2+75b^2a
= -3^5 * ac^2 + 3*25*ab^2
= 3a(-3^4*c^2 + 25b^2)
= 3a((5b)^2 - (9c)^2)
= 3a(5b+9c)(5b-9c)

(2) p^2-(p+q)^2
= p^2 - p^2 - 2pq - q^2
= - 2pq - q^2
= -q(2p + q)

(3) 25(x+1)^2-16(x-1)^2
= [5(x+1)]^2 - [4(x-1)]^2
= [5(x+1) + 4(x-1)][5(x+1) - 4(x-1)]
= (5x + 5 + 4x - 4)(5x + 5 - 4x + 4)
= (9x + 1)(x + 9)