## Solutions to identities-I

In this page, 'Solutions to identities-I' we are discussing how to do the problems given in problems on algebraic identities.

### Solutions to identities-I

1.     Expand the following

(i)     (5x + 2y + 3z)²

Solution:

Here we have to use the identity

(x+y+z)²  =  x² + y² + z² + 2xy + 2yz + 2zx

(5x + 2y + 3z)²     = (5x)²+(2y)²+(3z)²+2(5x)(2y)+2(2y)(3z)+2(3z)(5x)

=  25x² + 4y² + 9z² + 20xy  + 12yz + 30zx

(ii)      (2a + 3b - c)²

Solution:

Here we have to use the identity

(x+y+z)²  =  x² + y² + z² + 2xy + 2yz + 2zx

(2a + 3b - c)² = (2a)²+(3b)²+(-c)²+2(2a)(3b)+2(3b)(-c)+2(-c)(2a)

=  4a² + 9b² + c² +12ab - 6bc - 4ca

(iii)     (x-2y-4z )²

Solution:

Here we have to use the identity

(x+y+z)²  =  x² + y² + z² + 2xy + 2yz + 2zx

(x-2y-4z )²    =  x² + (-2y)² + (-4z)² + 2(x)(-2y) + 2(-2y)(-4z) +

2 (-4z)(x)

=  x² + 4y² + 16z² - 4xy + 16yz - 8yz

(iv)     (p - 2q + r)²

Solution:

Here we have to use the identitiy

(x+y+z)²  =  x² + y² + z² + 2xy + 2yz + 2zx

(p - 2q + r)² =  (p)² +(-2q)² +(r)²+2(p)(-2q)+2(-2q)(r)+2(r)(p)

=   p² +  4q² + r² - pq - 4qr + 2rp

2.      Find the expansion of

(i)     (x+1)(x+4)(x+7)

Solution:

Here we have to use the identity

(x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ca)x + abc

(x+1)(x+4)(x+7) = x³ + (1+4+7)x² + (1.4+4.7+7.1)x + 1.4.7

=  x³ +  12x² + (4+28+7)x + 28

=  x³ + 12x² + 39x + 28

(ii)    (p+2)(p-4)(p+6)

Solution:

Here we have to use the identity

(x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ca)x + abc

(p+2)(p-4)(p+6) = p³ + (2-4+6)p² +(2(-4)+(-4)6+6(2))p + (2)(-4)(6)

= p³ + 4p² + (-8-24+12)p-48

=  p³ + 4p² - 20p - 48

(iii)    (x+5)(x-3)(x-1)

Solution:

Here we have to use the identity

(x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ca)x + abc

(x+5)(x-3)(x-1)  = x³ + (5-3-1)x²  + (5(-3)+(-3)(-1)(-1)5)x+5(-3)(-1)

= x³ + x² + (-15+3-5)x + 15

= x³ + x²- 17x + 15

(iv)    (x-a)(x-2a)(x-4a)

Solution:

Here we have to use the identity

(x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ca)x + abc

(x-a)(x-2a)(x-4a)= x³ + (-a-2a-3a)x² + [(-a)(-2a)+(-2a)(-3a)+(-3a)                                                           (-a)]x + (-a)(-2a)(-3a)

= x³-6ax² + (2a²+6a²+3a²)x-6a³

= x³-6ax² + 11a²x - 6a³

(v)     (3x+1)(3x+2)((3x+5)

Solution:

Here we have to use the identity

(x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ca)x + abc

(3x+1)(3x+2)((3x+5)  = (3x)³ + (1+2+5)(3x)² + (1.2+2.5+5.1)(3x)+1.2.5

= 27 x³ + (8)9x² + (2+10+5)(3x)+1.2.5

= 27 x³ + 72x² + 51x+ 10

(vi)     (2x+3)(2x-5)(2x-7)

Solution:

Here we have to use the identity

(x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ca)x + abc

(2x+3)(2x-5)(2x-7)       = (2x)³ + (3-5-7)(2x)² + [(3)(-5)+(-5)(-7)+(-7)(3)]                                                          (2x) + (3)(-5)(-7)

=  8x³ + (-9)(4x²) + (-15+35-21)(2x) + 105

=  8x³ - 36 x² - 2x + 105

Students can solve the problems on their own, compare the answer with the solutions discussed above in'Solutions to identities-I'. If you are having any doubt you can contact us through mail, we will help you to clear your doubts.