Question

Ex 12,13 !!!!!!!!!!!!

Answer 1

12) a)x²-8=x²-(2√2)²=(x-2√2)(x+2√2)
b)12-y²=(2√3)²-y²=(2√3-y)(2√3+y)
c)a²-2=a²-(√2)²=(a-√2)(a+√2)
d)5-b²=(√5)²-b²=(√5-b)(√5+b)
e)4x²-15=(2x)²-(√15)²=(2x-√15)(2x+√15)
f)12x²-y²=(2√3x)²-y²=(2√3x-y)(2√3x+y)
g)$3x ^{2} - \frac{2}{9} y^{2} =( \sqrt{3}x) ^{2}- (\frac{ \sqrt{2} }{3}y) ^{2}=( \sqrt{3}x- \frac{ \sqrt{2} }{3})( \sqrt{3}x+ \frac{ \sqrt{2} }{3})$
h)$\frac{11}{4}x ^{2}-14=( \frac{ \sqrt{11} }{2}x) ^{2}- \sqrt{14}=( \frac{ \sqrt{11} }{2}x- \sqrt{14})(\frac{ \sqrt{11} }{2}x+ \sqrt{14})$
i)$\frac{1}{9}-5 x^{2} = (\frac{1}{3}) ^{2}- ( \sqrt5}x) ^{2}=( \frac{1}{3}- \sqrt{5}x)( \frac{1}{3}+ \sqrt{5}x)$
j)$( \frac{2}{5}) ^{2}-( \sqrt{7} x)^{2}=( \frac{2}{5}- \sqrt{7}x)(\frac{2}{5}+ \sqrt{7}x)$
k)$(\frac{3}{7}) ^{2}- ( \sqrt{2}x) ^{2}=( \frac{3}{7}- \sqrt{2}x)(\frac{3}{7}+ \sqrt{2}x)$
l)$(\sqrt{7}x) ^{2}- (\frac{1}{5}) ^{2}=( \sqrt{7}x- \frac{1}{5})( \sqrt{7}x+ \frac{1}{5})$
m)$(2 \sqrt{2}x) ^{2} - (\frac{y}{7}) ^{2}=(2 \sqrt{2}x- \frac{y}{7})(2 \sqrt{2}x+ \frac{y}{7})$
13)a) [tex] (\frac{1}{3a}x )^{2}-(2y) ^{2}=( \frac{1}{3a}x-2y)(\frac{1}{3a}x+2y) [/tex]
b)$(\frac{5x}{6y}) ^{2}- (7a)^{2}=(\frac{5x}{6y}-7a) (\frac{5x}{6y}+7a)$