Question 33091
a)x^2-3x-10
x^2-5x+2x-10
x(x-5)+2(x-5)
(x+2)(x-5)
b)x^2-x-20
X^2-5x+4x-20
x(x-5)+4(x-5)
(x+4)(x-5)
c)(2x^2+5x-12) /( 2x^2-7x+6)
(2x^2+8x-3x-12)/(2x^2-4x-3x+6)
{2x(x+4)-3(x+4)}/{2x(x-2)-3(x-2)}
{(2x-3)(x+4)}/{(2x-3)(x-2)}
cancelling (2x-3)on both numerator and denominator we get
(x+4)/(x-2)
d)9x^2-16 / 3x^2-x-4 
{(3x)^2 - (4)^2}/(3x^2 - 4x-+3x -4)
{(3x-4)(3x+4)}/{x(3x-4)+(3x-4)} numerator: according to a^2 - b^2 = (a+b)*(a-b)
{(3x-4)(3x+4)}/{(x+1)(3x-4)}
cancelling (3x-4) on both numerator and denominator we get
(3x+4)/(x-1)
e)square root[144x^10y^12z^18] 
{{{sqrt((12)^2(x^5)^2(y^6)^2(z^9)^2))}}}
{{{12(x^5)(y^6)(z^9)}}}
f)square root[50]+2 square root[32]- 2 square root[8] 
{{{sqrt(50)}}}+2{{{sqrt(32)}}}-2{{{sqrt(8)}}}
{{{sqrt(25*2)}}}+2{{{sqrt(16*2)}}}-2{{{sqrt(4*2)}}}
{{{5*sqrt(2)}}}+{{{2*4(sqrt(2))}}}-{{{2*2(sqrt(2))}}}
{{{sqrt(2)*(5+8-4)}}}
{{{9*(sqrt(2))}}}
g) Rationalize the denominator:
 
 2/{{{(sqrt(6)) -(sqrt(5))}}}
rationalising the denominator by multipling with {{{(sqrt(6)) +(sqrt(5))}}}
2 * [{{{(sqrt(6)) +(sqrt(5))}}}]/{{{sqrt(6)^2 -(sqrt(5)^2)}}}
2 * [{{{(sqrt(6)) +(sqrt(5))}}}]