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For Class IX (Science Group)
Mathematics I
Supplementary Examination 2025
Special Thanks To Sir Kashif Raza
Solution
Section : "C"
Q.12: FACTORIZATION:(v) a6 + 1
Solution:
⇒ a6 + 1
⇒ {(a2)3 + (1)3}
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2) ]
⇒ (a2 +1)[(a2)2 - (a2)(1) + (1)2]
⇒ (a2 + 1)(a4 - a2 + 1
Therefore,
⇒ a6 + 1 = (a2 + 1)(a4 - a2 + 1 ) Ans
(vi) x4 + x2 + 25
Solution:
⇒ (x4 + 25) + x2 (Rearrange the terms)
By adding & subtracting 2(x2)(5) = 10x2, we get:
⇒ (x4 + 10x2 + 25) - 10x2y2 + x2
By using formula: [∵ a2 + 2ab + b2 = (a + b)2] a perfect square
⇒ {(x2)2 + 2(x2)(5) + (5)2 - 9x2y2
⇒ (x2 + 5)2 - 9x2
Now By using formula [∵ a2 - b2 = (a - b)(a + b)] difference of two squares
⇒ (x2 + 5)2 - (3x)2
⇒ (x2 + 5 - 3x)(x2 + 5 + 3x)
⇒ (x2 - 3x + 5)(x2 + 3x + 5) Ans
no tyari😊
ReplyDeleteAssalamualikum
ReplyDeleteSir new syllabus schedule agaya he BSEK ki taraf se
so plz update it
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We have updated revised date sheet
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DeleteAssalmualikum miss
ReplyDeletea⁶ + 1 me a³ + b³ apply hoga or phr ( a+b)² ya to phr a³ + b³ pe hi end hogay ga Or x + xy +25 me middle term breaking apply hoga ??
Wa alaikum Assalam, We have uploaded solution.
DeleteFurther more
Factorization k total six sequence formula hain
1. check common
2. perfect square (a^2 土 2ab + b^2)
3. difference of square (a^2 - b^2)
4. breaking method
5. Cube a^3 +3a^2b+3ab^2 +b^3 = (a+b)^3 OR a^3 -3a^2b+3ab^2 -b^3 = (a-b)^3
6. Cube (a^3 土 b^3) = (a土b)(a^2土ab+b^2)
so according to it
a⁶ + 1 main 6th formula use hoga
jab k
x4 + x2 + 25 main do formula use hogain
formula 2 : perfect square by adding and and subtracting middle value than formula 3 difference of square.
mid term braeking se solve ho jaiye ga lakin but incorrect proper method oper diya hain
JAZAKALLAH