Factorize completely \((x^2 + x)^2 - (2x + 2)^2\)
(x + 1)(x + 2)(x - 2)
(x + 1) 2(x + 2) (x - 2)
(x + 1)2 (x + 2)2
(x + 1)2 (x + 2)(x - 2)
Correct answer is D
\((x^{2} + x)^{2} - (2x + 2)^{2}\)
= \((x^{2} + x + 2x + 2)(x^{2} + x - (2x + 2))\)
= \((x^{2} + 3x + 2)(x^{2} - x - 2)\)
= \(((x + 1)(x + 2))((x + 1)(x - 2))\)
= \((x + 1)^{2} (x + 2)(x - 2)\)
If (g(y)) = \(\frac{y - 3}{11}\) + \(\frac{11}{y^2 - 9}\). what is g(y + 3)?
\(\frac{y}{11} + \frac{11}{y(y + 6)}\)
\(\frac{y}{11} + \frac{11}{y(y + 3)}\)
\(\frac{y + 30}{11} + \frac{11}{y(y + 3)}\)
\(\frac{y + 3}{11} + \frac{11}{y(y - 6)}\)
Correct answer is A
\(g(y) = \frac{y - 3}{11} + \frac{11}{y^{2} - 9}\)
\(\therefore g(y + 3) = \frac{(y + 3) - 3}{11} + \frac{11}{(y + 3)^{2} - 9}\)
\(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y + 9 - 9}\)
\(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y}\)
= \(\frac{y}{11} + \frac{11}{y(y + 6)}\)
If a = -3, b = 2, c = 4, evaluate \(\frac{a^3 - b^3 - c^{\frac{1}{2}}}{b - a - c}\)
37
\(\frac{-37}{5}\)
\(\frac{37}{5}\)
-37
Correct answer is D
\(\frac{a^{3} - b^{3} - c^{\frac{1}{2}}}{b - a - c} = \frac{(-3)^{3} - (2)^{3} - 4^{\frac{1}{2}}}{2 - (-3) - 4}\)
= \(\frac{-37}{1} = -37\)
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
\(\frac{1}{3}\)
\(\frac{2}{3}\)
\(\frac{8}{27}\)
\(\frac{4}{9}\)
Correct answer is C
\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)
When y = 8, x = 1
\(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\)
\(k = 2\)
\(\therefore x = \frac{2}{\sqrt[3]{y}}\)
When x = 3,
\(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\)
\(y = (\frac{2}{3})^{3} = \frac{8}{27}\)
\(\frac{a + b}{a - b}\)
\(\frac{1}{a^2 - b^2}\)
\(\frac{a - b}{a + b}\)
a2 - b2
Correct answer is B
\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\)
\(\frac{1}{p} = \frac{(a + b)^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{(a - b)^{2}}\)
\(\therefore p = \frac{a - b}{(a + b)^{2}}\)
\(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\)
= \(\frac{1}{(a + b)(a - b)}\)
= \(\frac{1}{a^{2} - b^{2}}\)