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Comment on the expressions obtained in (iii) and (iv).

Mathematics

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Answer

iv

4x2+4x+14x^2 + 4x + 1

Step-by-step solution

3 steps

Part iv· 3 steps

Expand (2x+1)2(2x+1)^2.

  1. Tumia fomula ya kupanua binomial iliyopigwa mraba

    (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. Hapa, a=2xa=2x na b=1b=1.

    (2x+1)2=(2x)2+2(2x)(1)+(1)2(2x+1)^2 = (2x)^2 + 2(2x)(1) + (1)^2
  2. Rahisisha kila neno.

    (2x)2=4x2(2x)^2 = 4x^2 2(2x)(1)=4x2(2x)(1) = 4x (1)2=1(1)^2 = 1
  3. Unganisha maneno yaliyopatikana.

    (2x+1)2=4x2+4x+1(2x+1)^2 = 4x^2 + 4x + 1

    Kupanua (2x+1)2(2x+1)^2 kunatoa 4x2+4x+1\boxed{4x^2 + 4x + 1}.

Part v

Comment on the expressions obtained in (iii) and (iv).

Katika sehemu (iii), eneo la mraba lenye pande (2x+1)(2x+1) units lilipatikana kuwa 4x2+4x+14x^2 + 4x + 1.
Katika sehemu (iv), tulipanua usemi (2x+1)2(2x+1)^2 na tukapata 4x2+4x+14x^2 + 4x + 1.

Maoni:
Maneno yaliyopatikana katika sehemu (iii) na (iv) ni sawa. Hii inathibitisha kwamba eneo la mraba lenye urefu wa upande (2x+1)(2x+1) units linaweza kuwakilishwa kama (2x+1)2(2x+1)^2, na kupanua usemi huu kunatoa eneo la 4x2+4x+14x^2 + 4x + 1. Hii inaonyesha uhusiano kati ya jiometri (eneo la mraba) na aljebra (upanuzi wa binomial).

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