Expand each of the following, using suitable identities: [3 MARKS]
1)(x+2y+4z)2
2)(3x−7y−z)2
Formula: 1 Mark
Subsitution: 0.5 Mark each
Calculation: 0.5 Mark each
(x+2y+4z)2
=x2+(2y)2+(4z)2+2(x)(2y)+2(2y)(4z)+2(4z)(x)
=(x)2+4(y)2+16(z)2+4xy+16yz+8zx
2)(3x−7y−z)2
a=3x,b=−7y,c=−z
Substituting the values of a, b and c in (1)
(3x−7y−z)2
=(3x)2+(−7y)2+(−z)2+2(3x)(−7y)+2(−7y)(−z)+2(−z)(3x)
=9x2+49y2+z2−42xy+14yz−6xz