Form 2 Maths

2009-02-27 4:50 am
(a) Expand (5-2x)(x+4)
and the answer is - 2x^2 - 3x + 20
(b) From the result of (a), simplify
(5-2x)(x+4) - (2x+7)
and the answer is - 2x^2 - 5x - 13
(c) If the constant term of (5-2x)(x+4) - (2x+7) + k is equal to 10, find the value of k.
Please help me (c) part and show clear steps and check whether the answers in (a) and (b) are correct. Thanks for help!
更新1:

001: In part (c), your answer is incorrect since the question have been said that "find the value of k"

回答 (3)

2009-02-27 5:57 am
✔ 最佳答案
for (b) it should be:

(5-2x)(x+4) - (2x+7)
=5x+20-2x^2-8x-2x-7
=-2x^2-3x+13

(c)

(5-2x)(x+4) - (2x+7)+k
=-2x^2-3x+(13+k)

the constant term=13+k=10
therefore k= -3
2009-02-27 7:42 pm
We only know the range of k.
2009-02-27 5:03 am
(a)
(5-2x)(x+4)
=5(x+4)-2x(x+4)
=5x+20-2x^2-8x
= - 2x^2 - 3x + 20
(b)
(5-2x)(x+4) - (2x+7)
=20-2x^2-3x-2x-7
= - 2x^2 - 5x - 13
(c)
(5-2x)(x+4) - (2x+7) + k=10
-2x^2 - 5x - 13+k=10
-2x^2 - 5x+k=23
k=23+2x^2+5x




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