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.

Solution:

a2 + 2ab + b2= (a+b)2

2ab = (a+b)2 – (a2 + b2) = (2)2 – (8) = 4 – 8 = – 4. ∴ ab = 21(2ab) = 21(–4) = – 2.

a3+ b3 = (a+b)3 – 3ab(a+b) = (2)3 – 3(–2)(2) = 8 – 3 (– 4) = 8 + 12 = 20.

Problem-5 . Factorize (x – y)

^{3 }+ (y – z)^{3 }+ (z – x)^{3}.Solution: Put X = x – y, Y = y – z, Z = z – x. Then

X + Y + Z = (x – y) + (y – z) + (z – x) = x – y + y – z + z – x = 0.

∴ X

^{3 }+ Y^{3 }+ Z^{3 }= 3XYZSubstituting for X, Y and Z

(x – y)

^{3 }+ (y – z)^{3 }+ (z –x)^{3 }= 3(x – y) (y – z) (z – x).Problem-6. Factorize x

^{4 }+1.Solution: Adding and subtracting 2x

^{2}, the given expression isx

^{4 }+ 1 = (x^{4 }+ 2x^{2 }+ 1) – 2x^{2 }= [(x

^{2})^{2 }+ 2(x^{2})(1) + (1)^{2}] – 2x^{2 }= (x

^{2 }+ 1)^{2 }– ( 2x)^{2 }= [(x

^{2 }+ 1) + (2x)][(x^{2 }+ 1) – (2x)]= (x

^{2 }+ 2x + 1) (x^{2 }– 2x + 1).Problem-7 Factorize x

^{4 }+ x^{2}y^{2 }+ y^{4}.Solution: Adding and subtracting x

^{2}y^{2}, we getx

^{4 }+ x^{2}y^{2 }+ y^{4 }= (x^{4 }+ 2x^{2}y^{2 }+ y^{4}) – x^{2}y^{2 }= [(x

^{2})^{2 }+ 2(x^{2})(y^{2}) + (y^{2})^{2}] – (xy)^{2 }= (x

^{2 }+ y^{2})^{2 }– (xy)^{2 }= [x

^{2 }+ y^{2 }+ (xy)] [x^{2}+y^{2 }– (xy)]= (x

^{2 }+ xy + y^{2}) (x^{2 }– xy + y^{2}).Problem-8 Factorize x

^{4 }+ 5x^{2 }+ 9.Solution : Adding and subtracting x

^{2}, we getx

^{4 }+ 5x^{2 }+ 9 = (x^{4 }+ 6x^{2 }+ 9) – x^{2 }= (x

^{2 }+ 3)^{2 }– x^{2 }= [(x^{2 }+ 3) + x][(x^{2}+3) – x]= (x

^{2 }+ x + 3) (x^{2 }– x + 3).Do yourself:

1. If a + b = 5 and a – b = 4, find a

^{2 }+ b^{2 }and ab.2. If a + b = 10 and ab = 20, find a

^{2 }+ b^{2 }and (a–b)^{2}.3. If (x + l)(x + m) = x

^{2 }+ 4x + 2, find l^{2 }+ m^{2 }and (l – m)^{2}.4. If a + b + c = 11 and ab + bc + ca = 38, find a

^{2 }+ b^{2 }+ c^{2}.5. If (x + a)(x + b)(x + c) ≡ x

^{3 }– 9x^{2 }+ 23x – 15, find a + b + c, 1/a + 1/b + 1/c and a^{2 }+ b^{2 }+ c^{2}6. If 2a – 3b = 2 and ab = 6, find 8a

^{3 }– 27b^{3}.7. If ,31=+xx find x

^{2 }+21x and x^{3}+ 31x.8. . If x + y = 6 and xy = 8, find x

^{2 }+ y^{2 }and x^{3 }+ y^{3}.9. . If p + q = 6 and p

^{2 }+ q^{2 }= 32, find pq, p^{3 }+ q^{3 }and p^{4 }+ q^{4}.Factorization formulae

(ii) (X – Y)

^{2 }= X^{2 }– 2XY + Y^{2 }(iii) (X + Y)(X–Y) = X

^{2 }– Y^{2 }(iv) ( X + Y) (X

^{2 }– XY + Y^{2}) = X^{3 }+ Y^{3 }(v) (X – Y)(X

^{2 }+ XY + Y^{2}) = X^{3 }– Y^{3 }(vi) (X + Y)

^{3 }= X^{3 }+ Y^{3 }+ 3X^{2}Y + 3XY^{2 }= X^{3 }+ Y^{3 }+ 3X Y (X +Y )(vii) (X – Y)

^{3 }= X^{3 }– Y^{3 }– 3X^{2}Y + 3XY^{2 }= X^{3 }– Y^{3 }– 3X Y (X –Y )(viii) (X + Y + Z)

^{2 }= X^{2 }+ Y^{2 }+ Z^{2 }+ 2XY + 2YZ + 2ZX(ix) (X + Y + Z) (X

^{2 }+ Y^{2 }+Z^{2 }– XY – YZ – ZX) = X^{3 }+ Y^{3 }+ Z^{3 }– 3XYZFactorize the polynomial using the factorization formulae:

10. 1 + 6x + 9x

^{2 }11. 144x

^{2 }– 72x + 912. 4a

^{2}b^{2 }+ 20abcd + 25c^{2}d^{2 }13. x

^{2 }+ y^{2 }– a^{2 }– b^{2 }+ 2xy + 2ab14. 33x

^{3}y^{3 }+ 27z^{3 }15. (x + y)

^{3 }+ 8y^{3}16. (x

^{2 }+ 1)^{3 }+ (x^{2 }– 1)^{3 }17. x

^{6 }– y^{6 }14. (x+y)

^{3 }− (x–y)^{3 }15. (p+q)

^{3 }+ (p–q)^{3 }+ 6p (p^{2 }– q^{2})16. 27x

^{3 }+ y^{3 }+ 27x^{2}y + 9xy^{2}^{ }17. x

^{3 }– 12x

^{2 }+48x – 64

18. 8x

^{3 }– 27y^{3 }– 36x^{2}y + 54xy^{2 }19. 4x

^{2 }+ 9y^{2 }+ z^{2 }+ 12xy + 4xz + 6yz20. a

^{2 }+ b^{2 }+ 9c^{2 }+ 2ab – 6ac – 6bc21. x

^{3 }– y^{3 }+ 1 + 3xy22. 8x

^{3 }– 125y^{3 }+ 180xy +21623. 8x

^{3 }– 27y^{3 }+z^{3}+18xyz24. 33a

^{3 }– 8b^{3 }– 125c^{3 }− 30 3abc25. (a –2b)

^{3 }+ (2b – 3c)^{3 }+ (3c – a)^{3}26. (x + y – 2z)

^{3 }+ (y + z – 2x)^{3 }+ (z + x – 2y)^{3}27. (a

^{2 }– b^{2})^{3 }+ (b^{2 }– c^{2})^{3 }+ (c^{2 }– a^{2})^{3 }28. a

^{3}(b – c)^{3 }+ b^{3}(c – a)^{3 }+ c^{3}(a – b)^{3} 29. x(x + z) – y(y + z)

30. 1–2xy–x

^{2 }– y^{2} 31. x

^{4 }+4**Factorization of the quadratic expression ax**

^{2 }**+ bx + c**

Resolve into factors each of the following:

1. x

^{2 }+ 7x +122. x

^{2 }+ 9x + 203. d

^{2 }+ 10d +214. z

^{2 }– 7z – 985. a

^{2 }– a –726. x

^{2 }+ x – 907. p

^{2 }– 8p +158. y

^{2 }– 13y + 429. y

^{2 }– 20y + 9910. t

^{2 }– 28t + 195Factorize each of the following:

11. 2a

^{2 }+ 13a + 1512. 4x

^{2 }+ 8x + 313. 4x

^{2 }+ 12x + 914. 6x

^{2 }+ x – 115. 6p

^{2 }+ 17p + 1016. 4a

^{2 }– 11a – 1517. 7m

^{2 }+ 16m – 1518. 8p

^{2 }+ 29p – 1219. 6x

^{2 }+ 5x – 620. 15y

^{2 }– 13y – 621. 14x

^{2 }– x – 322. 9a

^{2 }– 9a + 223. 2a

^{2 }– 13a + 1824. 12x

^{2 }– 7x + 125. 16x

^{2 }– 32x +7Resolve into factors each of the following :

26. 9x

^{2 }+ 24xy + 15y^{2 }27. 4x

^{2 }– 16xy – 9y^{2 }28. 6c

^{2 }+ 11cd – 10d^{2 }29. 5x

^{2 }– 11xy + 6y^{2 }30. 2a

^{2 }– 15ab + 28b^{2 }Factorize the following:

31. √2 x

^{2 }+ 3x +√232. √3 x

^{2}+ 11x + 6√333. 5√5 x

^{2}+ 20 x + 3 √534. 2x

^{2}+ 3√5 x + 535. 7x

^{2}+ 2√14 x +2
## No comments:

## Post a Comment