Question

(x+1) (x+2) (3x-1) (3x-4)+12

Answer 1

$= ({x }^{2} + 3x + 2)(9 {x}^{2} - 15x + 4) + 12 \\ = ( {x}^{2} + 3x + 2) (1)(9 - 5 + 2) + 12 \\ = ({x}^{2} + 3x + 2)(6) + 12 \\ = 6 {x}^{2} + 18x + 24$

Answer 2

GIVEN :

The expression is (x+1) (x+2) (3x-1) (3x-4)+12

TO FIND :

The product of the given expression.

SOLUTION :

(x+1) (x+2) (3x-1) (3x-4)+12=[(x+1) (x+2)][ (3x-1) (3x-4)]+12

By using the Distributive Property :

(a+b)(x+y)=a(x+y)+b(x+y)

=[x (x+2)+1(x+2)][ 3x(3x-4)-1(3x-4)]+12

=[x(x)+x(2)+1(x)+1(2)][3x(3x)+3x(-4)-1(3x)-1(-4)]+12

By using the identity :

$a^m.a^n=a^{m+n}$

$=[x^{1+1}+x+2][9x^{1+1}-12x-3x+4]+12$

$=[x^2+x+2][9x^2-12x-3x+4]+12$

Adding the like terms

$=[x^2+x+2][9x^2-15x+4]+12$

By using the Distributive Property :

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

$=(x^2)(9x^2-15x+4)+x[9x^2-15x+4)+2(9x^2-15x+4)+12$

$=x^2(9x^2)+x^2(-15x)+x^2(4)+x(9x^2)+x(-15x)+x(4)+2(9x^2)+2(-15x)+2(4)+12$

By using the identity :

$a^m.a^n=a^{m+n}$

$=9x^{2+2}-15x^{2+1}+4x^2+9x^{1+2}-15x^{1+1}+4x+18x^2-30x+8+12$

$=9x^4-15x^3+4x^2+9x^3-15x^2+4x+18x^2-30x+8+12$

By adding the like terms

$=9x^4+12x^3-23x^2-18x+20$

∴ $(x+1) (x+2) (3x-1) (3x-4)+12=9x^4+12x^3-23x^2-18x+20$

Therefore the product of the given expression (x+1) (x+2) (3x-1) (3x-4)+12 is $9x^4+12x^3-23x^2-18x+20$