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 a

^{2 }+ b^{2 }and (a + b)^{2}.Solution: a

^{2 }+ b^{2 }= (a – b)^{2 }+ 2ab = 6^{2 }+ 2(40) = 36 + 80 = 116 (a + b)

^{2 }= (a – b)^{2 }+ 4ab = 6^{2 }+ 4(40)= 36 + 160 = 196Problem-2. If (x + p)(x + q) = x

^{2 }– 5x – 300, find the value of p^{2 }+ q^{2}.Solution:

By product formula, we have (x + p) (x + q) = x

^{2 }+ (p + q)x + pq.So, by comparison, we get p + q = –5, pq = –300.

Now, we have p

^{2 }+ q^{2 }= (p + q)^{2 }– 2 pq = (–5)^{2 }–2(–300) = 25 + 600 = 625.Problem-3. If (x + a)(x + b)(x + c) ≡ x

^{3 }– 6x^{2 }+ 11x – 6, find the value of a^{2 }+ b^{2}+ c^{2}.Solution: From the product formula, we have

(x+a)(x+b)(x+c) = x

^{3 }+ (a + b + c)x^{2 }+ (ab + bc + ca)x + abc.Comparing, we get a + b + c = –6, ab + bc + ca = 11, abc = –6.

∴ a

^{2 }+ b^{2 }+ c^{2 }= (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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