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Problem related to factorization,ratiolization

By Jsunil
CBSE MATH STUDY: Problem related to factorization,ratiolization:Problem-1 If the values of a – b and ab are 6 and 40 respectively, find the values of a2 + b2 and (a + b)2.
Solution: a2 + b2 = (a – b)2 + 2ab = 62 + 2(40) = 36 + 80 = 116
(a + b)2 = (a – b)2 + 4ab = 62 + 4(40)= 36 + 160 = 196
Problem-2. If (x + p)(x + q) = x2 – 5x – 300, find the value of p2 + q2.
Solution:
By product formula, we have (x + p) (x + q) = x2 + (p + q)x + pq.
So, by comparison, we get p + q = –5, pq = –300.
Now, we have p2 + q2 = (p + q)2 – 2 pq = (–5)2 –2(–300) = 25 + 600 = 625.
Problem-3. If (x + a)(x + b)(x + c) ≡ x3 – 6x2 + 11x – 6, find the value of a2 + b2+ c2.
Solution: From the product formula, we have
(x+a)(x+b)(x+c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc.
Comparing, we get a + b + c = –6, ab + bc + ca = 11, abc = –6.
∴ a2 + b2 + c2 = (a + b + c)2 –2 (ab + bc + ca) = (– 6)2 – 2(11) = 36 – 22 = 14.
Problem-4 If a+b=2 and a2+b2=8,find a3+b3 and a4+b4.
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