Problem related to factorization,ratiolization

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+ b²and (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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