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Question
Simplify:
\[\frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
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Solution
It is known that,
a2 − b2 = (a + b) (a − b)
\[\ \frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
\[ = \frac{4 x^2 - 8x - 3x + 6}{\left(4x \right)^2 - \left(3 \right)^2}\]
\[ = \frac{4x\left(x - 2 \right) - 3\left(x - 2 \right)}{\left(4x + 3 \right)\left(4x - 3 \right)}\]
\[ = \frac{\left(4x - 3 \right)\left(x - 2 \right)}{\left(4x + 3 \right)\left( 4x - 3 \right)}\]
\[ = \frac{x - 2}{4x + 3}\]
RELATED QUESTIONS
Simplify:
\[\frac{m^2 - n^2}{\left( m + n \right)^2} \times \frac{m^2 + mn + n^2}{m^3 - n^3}\]
Simplify:
\[\frac{3 x^2 - x - 2}{x^2 - 7x + 12} \div \frac{3 x^2 - 7x - 6}{x^2 - 4}\]
Simplify:
\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
Factorise:
y3 − 27
Factorise:
x3 − 64y3
Factorise:
8p3 −\[\frac{27}{p^3}\]
Factorise:
`16a^3 - 128/b^3`
Simplify:
(3a + 5b)3 − (3a − 5b)3
Factorise: `a^3 - 1/(a^3)`
Factorise the following:
27x3 – 8y3
