F. 2 Maths (最佳解答取30分)_0－gbu94mz95e的部落格

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F. 2 Maths (最佳解答取30分)

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Factorization by using identities(Difference of two squares and Perfect square)Factorize the following expressions.1) -243ac^2+75b^2a2) p^2-(p+q)^23) 25(x+1)^2-16(x-1)^2Remarks:1) Please show the working steps clearly.(3 points)2) Explain/state the working steps in details. (2 points)3) Make sure... 顯示更多 Factorization by using identities (Difference of two squares and Perfect square) Factorize the following expressions. 1) -243ac^2+75b^2a 2) p^2-(p+q)^2 3) 25(x+1)^2-16(x-1)^2 Remarks: 1) Please show the working steps clearly.(3 points) 2) Explain/state the working steps in details. (2 points) 3) Make sure your answer is correct. (Total points of each question =5 points) -->5 X 3 X 2=30 points Thanks a lot!

最佳解答:

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)

其他解答:

1) -243ac^2+75b^2a ans: -243ac^2+75b^2a =-3a(81c^2-25b^2) 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)