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Question: Factor (41-43). 41. $n^4+4$. 42. $1+n^4+n^8$. 43. $1+x^5$, Simplify the following expressions (44-16...

Factor (41-43). 41. n4+4n^4+4. 42. 1+n4+n81+n^4+n^8. 43. 1+x51+x^5, Simplify the following expressions (44-166). 44. a2−b2a−b−a3−b3a2−b2\frac{a^2-b^2}{a-b}-\frac{a^3-b^3}{a^2-b^2}. 45. 1(a−b)(a−c)+1(b−c)(b−a)+1(c−a)(c−b)\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}. 46. (a+aba−b)(aba+b−a):a2+b2a2−b2(a+\frac{ab}{a-b})(\frac{ab}{a+b}-a):\frac{a^2+b^2}{a^2-b^2}. 47. (4(a+b)2ab−16)÷(a+b)2−abab:a3−b3ab(\frac{4(a+b)^2}{ab}-16)\div\frac{(a+b)^2-ab}{ab}:\frac{a^3-b^3}{ab}. 48. (a+3b(a−b)2+a−3ba2−b2):a2+3b2(a−b)2(\frac{a+3b}{(a-b)^2}+\frac{a-3b}{a^2-b^2}):\frac{a^2+3b^2}{(a-b)^2}. 49. (m+n−4mnm+n):(mm+n−nn−m−2mnm2−n2)(m+n-\frac{4mn}{m+n}):(\frac{m}{m+n}-\frac{n}{n-m}-\frac{2mn}{m^2-n^2}). 50. (12(11))(\frac{1}{2}(\frac{1}{1}))

Answer
  1. (n2−2n+2)(n2+2n+2)(n^2-2n+2)(n^2+2n+2)
  2. (n4+n2+1)(n4−n2+1)(n^4+n^2+1)(n^4-n^2+1)
  3. (1+x)(1−x+x2−x3+x4)(1+x)(1-x+x^2-x^3+x^4)
  4. aba+b\frac{ab}{a+b}
  5. 0
  6. −a4a2+b2-\frac{a^4}{a^2+b^2}
  7. 4ab(a−b)(a2+ab+b2)2\frac{4ab(a-b)}{(a^2+ab+b^2)^2}
  8. 2a+b\frac{2}{a+b}
  9. m−nm-n
  10. 12\frac{1}{2}
Explanation

Solution

Below are the step‐by‐step minimal solutions and final answers.

41. Factor n4+4n^4+4
• Write as a difference of two squares:

n4+4=(n4+4n2+4)−4n2=(n2+2)2−(2n)2.n^4+4 = (n^4+4n^2+4)-4n^2 = (n^2+2)^2 - (2n)^2.

• Factor using difference of squares:

(n2+2−2n)(n2+2+2n)=(n2−2n+2)(n2+2n+2).(n^2+2-2n)(n^2+2+2n) = (n^2-2n+2)(n^2+2n+2).

Answer 41: (n2−2n+2)(n2+2n+2)(n^2-2n+2)(n^2+2n+2)

42. Factor 1+n4+n81+n^4+n^8
• Substitute x=n2x=n^2 so that the expression becomes:

1+x2+x4.1+x^2+x^4.

• Recognize the factorization:

1+x2+x4=(x2+x+1)(x2−x+1).1+x^2+x^4 = (x^2+x+1)(x^2-x+1).

• Replace xx by n2n^2:

(n4+n2+1)(n4−n2+1).(n^4+n^2+1)(n^4-n^2+1).

Answer 42: (n4+n2+1)(n4−n2+1)(n^4+n^2+1)(n^4-n^2+1)

43. Factor 1+x51+x^5
• Use the formula for the sum of two fifth powers:

a5+b5=(a+b)(a4−a3b+a2b2−ab3+b4).a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4).

• Here, a=xa=x and b=1b=1:

1+x5=(1+x)(1−x+x2−x3+x4).1+x^5=(1+x)(1-x+x^2-x^3+x^4).

Answer 43: (1+x)(1−x+x2−x3+x4)(1+x)(1-x+x^2-x^3+x^4)

44. Simplify a2−b2a−b−a3−b3a2−b2\frac{a^2-b^2}{a-b}-\frac{a^3-b^3}{a^2-b^2}
• Notice that:
 a2−b2a−b=a+b\frac{a^2-b^2}{a-b}=a+b.
• Also, a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2) and a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b) so that

a3−b3a2−b2=a2+ab+b2a+b.\frac{a^3-b^3}{a^2-b^2}=\frac{a^2+ab+b^2}{a+b}.

• Then the expression becomes:

a+b−a2+ab+b2a+b=(a+b)2−(a2+ab+b2)a+b=aba+b.a+b - \frac{a^2+ab+b^2}{a+b}=\frac{(a+b)^2-(a^2+ab+b^2)}{a+b}=\frac{ab}{a+b}.

Answer 44: aba+b\displaystyle \frac{ab}{a+b}

45. Simplify 1(a−b)(a−c)+1(b−c)(b−a)+1(c−a)(c−b)\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}
• By symmetry (or by substituting particular values for a,b,ca,b,c) one verifies that the sum evaluates to 0.

Answer 45: 0

46. Simplify (a+aba−b)(aba+b−a):a2+b2a2−b2\Bigl(a+\frac{ab}{a-b}\Bigr)\Bigl(\frac{ab}{a+b}-a\Bigr):\frac{a^2+b^2}{a^2-b^2}
• Write the first bracket as:

a+aba−b=a(a−b)+aba−b=a2a−b.a+\frac{ab}{a-b}=\frac{a(a-b)+ab}{a-b}=\frac{a^2}{a-b}.

• Write the second bracket as:

aba+b−a=ab−a(a+b)a+b=−a2a+b.\frac{ab}{a+b}-a=\frac{ab-a(a+b)}{a+b}=\frac{-a^2}{a+b}.

• Their product is:

a2a−b⋅−a2a+b=−a4(a−b)(a+b)=−a4a2−b2.\frac{a^2}{a-b}\cdot\frac{-a^2}{a+b}=-\frac{a^4}{(a-b)(a+b)}=-\frac{a^4}{a^2-b^2}.

• Dividing by a2+b2a2−b2\frac{a^2+b^2}{a^2-b^2} gives:

−a4a2−b2⋅a2−b2a2+b2=−a4a2+b2.-\frac{a^4}{a^2-b^2}\cdot\frac{a^2-b^2}{a^2+b^2}=-\frac{a^4}{a^2+b^2}.

Answer 46: −a4a2+b2\displaystyle -\frac{a^4}{a^2+b^2}

47. Simplify (4(a+b)2ab−16)÷(a+b)2−abab:a3−b3ab\Bigl(\frac{4(a+b)^2}{ab}-16\Bigr)\div\frac{(a+b)^2-ab}{ab}:\frac{a^3-b^3}{ab}
• First simplify:

4(a+b)2ab−16=4(a+b)2−16abab=4[(a+b)2−4ab]ab=4(a−b)2ab.\frac{4(a+b)^2}{ab}-16=\frac{4(a+b)^2-16ab}{ab}= \frac{4[(a+b)^2-4ab]}{ab}= \frac{4(a-b)^2}{ab}.

• Then,

4(a−b)2ab(a+b)2−abab=4(a−b)2(a+b)2−ab.\frac{\frac{4(a-b)^2}{ab}}{\frac{(a+b)^2-ab}{ab}} =\frac{4(a-b)^2}{(a+b)^2-ab}.

• Note that:

(a+b)2−ab=a2+2ab+b2−ab=a2+ab+b2.(a+b)^2-ab=a^2+2ab+b^2-ab=a^2+ab+b^2.

• Next, divide by a3−b3ab\frac{a^3-b^3}{ab} with a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2):

4(a−b)2a2+ab+b2÷(a−b)(a2+ab+b2)ab=4(a−b)2a2+ab+b2⋅ab(a−b)(a2+ab+b2).\frac{4(a-b)^2}{a^2+ab+b^2}\div\frac{(a-b)(a^2+ab+b^2)}{ab}=\frac{4(a-b)^2}{a^2+ab+b^2}\cdot\frac{ab}{(a-b)(a^2+ab+b^2)}.

• Simplify to get:

4ab(a−b)(a2+ab+b2)2.\frac{4ab(a-b)}{(a^2+ab+b^2)^2}.

Answer 47: 4ab(a−b)(a2+ab+b2)2\displaystyle \frac{4ab(a-b)}{(a^2+ab+b^2)^2}

48. Simplify (a+3b(a−b)2+a−3ba2−b2):a2+3b2(a−b)2\Bigl(\frac{a+3b}{(a-b)^2}+\frac{a-3b}{a^2-b^2}\Bigr):\frac{a^2+3b^2}{(a-b)^2}
• Note that a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b). Express both terms of the sum with common denominator (a−b)2(a+b)(a-b)^2(a+b):
 – First term: a+3b(a−b)2=(a+3b)(a+b)(a−b)2(a+b)\dfrac{a+3b}{(a-b)^2}=\dfrac{(a+3b)(a+b)}{(a-b)^2(a+b)}.
 – Second term: a−3b(a−b)(a+b)=(a−3b)(a−b)(a−b)2(a+b)\dfrac{a-3b}{(a-b)(a+b)}=\dfrac{(a-3b)(a-b)}{(a-b)^2(a+b)}.
• Their sum:

(a+3b)(a+b)+(a−3b)(a−b)(a−b)2(a+b).\frac{(a+3b)(a+b)+(a-3b)(a-b)}{(a-b)^2(a+b)}.

• Expand:
 (a+3b)(a+b)=a2+4ab+3b2,(a−3b)(a−b)=a2−4ab+3b2.(a+3b)(a+b)=a^2+4ab+3b^2,\quad (a-3b)(a-b)=a^2-4ab+3b^2.
• Sum = 2a2+6b2=2(a2+3b2).2a^2+6b^2=2(a^2+3b^2).
• Thus the expression becomes:

2(a2+3b2)(a−b)2(a+b).\frac{2(a^2+3b^2)}{(a-b)^2(a+b)}.

• Now divide by a2+3b2(a−b)2\frac{a^2+3b^2}{(a-b)^2}:

2(a2+3b2)(a−b)2(a+b)⋅(a−b)2a2+3b2=2a+b.\frac{2(a^2+3b^2)}{(a-b)^2(a+b)}\cdot\frac{(a-b)^2}{a^2+3b^2}=\frac{2}{a+b}.

Answer 48: 2a+b\displaystyle \frac{2}{a+b}

49. Simplify (m+n−4mnm+n):(mm+n−nn−m−2mnm2−n2)(m+n-\frac{4mn}{m+n}):\Bigl(\frac{m}{m+n}-\frac{n}{n-m}-\frac{2mn}{m^2-n^2}\Bigr)
• Write the numerator over common denominator:

m+n−4mnm+n=(m+n)2−4mnm+n=m2−2mn+n2m+n=(m−n)2m+n.m+n-\frac{4mn}{m+n}=\frac{(m+n)^2-4mn}{m+n}=\frac{m^2-2mn+n^2}{m+n}=\frac{(m-n)^2}{m+n}.

• For the denominator, note:
 – nn−m=−nm−n \frac{n}{n-m}=-\frac{n}{m-n} and
 – m2−n2=(m−n)(m+n).m^2-n^2=(m-n)(m+n).
• Express each term with denominator (m+n)(m−n)(m+n)(m-n):
 $$ \frac{m}{m+n}=\frac{m(m-n)}{(m+n)(m-n)},\quad -\frac{n}{m-n}=\frac{-n(m+n)}{(m+n)(m-n)},\quad -\frac{2mn}{(m-n)(m+n)}.

•Combiningthenumerator:• Combining the numerator:

m(m-n)-n(m+n)-2mn=m^2-mn-nm-n^2-2mn=m^2-2mn-n^2=(m-n)^2.

• So the denominator equals $\frac{(m-n)^2}{(m+n)(m-n)}=\frac{m-n}{m+n}$. • Overall,

\frac{\frac{(m-n)^2}{m+n}}{\frac{m-n}{m+n}}=m-n.

**Answer 49:** $m-n$ --- **50. Simplify** $\frac{1}{2}\Bigl(\frac{1}{1}\Bigr)$ • It is simply: $\frac{1}{2}\cdot1=\frac{1}{2}$. **Answer 50:** $\displaystyle \frac{1}{2}$ --- **Metadata:** - **Subject:** Mathematics - **Chapter:** Polynomials (NCERT Class 12) - **Topic:** Factorization and Simplification - **Difficulty Level:** Medium - **Question Type:** multiple_choice (if taken from an exam paper) / descriptive (as steps are shown) Each solution uses standard algebraic manipulations suitable for a 12th grade student preparing for JEE/NEET.