If \((x + 2)\) and \((3x - 1)\) are factors of \(6x^{3} + x^{2} - 19x + 6\), find the third factor.
\(2x - 3\)
\(3x + 1\)
\(x - 2\)
\(3x + 2\)
Correct answer is A
To get the third factor, take the product of the other 2 factors and then divide the main equation by their product.
64.245
61.255
60.255
60.245
Correct answer is C
\((1.98)^{6} = (1 + 0.98)^{6} = 1 + 6(0.98) + 15(0.98)^{2} + 20(0.98)^{3} + 15(0.98)^{4} + 6(0.98)^{5} + (0.98)^{6}\)
\(\approxeq 1 + 5.88 + 14.406 + 18.823 + 13.836 + 5.424 + 0.886 \)
= \(60.255\)
Simplify \(\frac{1 + \sqrt{8}}{3 - \sqrt{2}}\)
\(7 + \sqrt{2}\)
\(7 + 7\sqrt{2}\)
\(1 - 7\sqrt{2}\)
\(1 + \sqrt{2}\)
Correct answer is D
\(\frac{1 + \sqrt{8}}{3 - \sqrt{2}}\)
Rationalizing by multiplying through with \(3 + \sqrt{2}\),
\((\frac{1 + \sqrt{8}}{3 - \sqrt{2}})(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}) = \frac{3 + \sqrt{2} + 3\sqrt{8} + 4}{9 - 2}\)
= \(\frac{3 + \sqrt{2} + 3\sqrt{4 \times 2} + 4}{7} \)
= \(\frac{7 + 7\sqrt{2}}{7} = 1 + \sqrt{2}\)
If \(8^{x} ÷ (\frac{1}{4})^{y} = 1\) and \(\log_{2}(x - 2y) = 1\), find the value of (x - y)
\(\frac{5}{4}\)
\(\frac{3}{5}\)
\(1\)
\(\frac{2}{3}\)
Correct answer is A
\(8^{x} ÷ (\frac{1}{4})^{y} = 1\)
\((2^{3})^{x} ÷ (2^{-2})^{y} = 2^{0}\)
\(2^{3x - (-2y)} = 2^{0}\)
\(\implies 3x + 2y = 0 .... (1)\)
\(\log_{2}(x - 2y) = 1\)
\( x - 2y = 2^{1} = 2 ..... (2)\)
Solving equations 1 and 2,
\(x = \frac{1}{2}, y = \frac{-3}{4}\)
\((x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}\)
If \(f(x) = 3x^{3} + 8x^{2} + 6x + k\) and \(f(2) = 1\), find the value of k.
-67
-61
61
67
Correct answer is A
\(f(x) = 3x^{3} + 8x^{2} + 6x + k\)
\(f(2) = 3(2^{3}) + 8(2^{2}) + 6(2) + k = 1\)
\(\implies 24 + 32 + 12 + k = 1\)
\(68 + k = 1 \therefore k = 1 - 68 = -67\)