Step 1: Start dividing by prime numbers
$156 \div 2 = 78$ (So one factor of 2)
$78 \div 2 = 39$ (Another factor of 2)
Step 2: Continue prime division
$39 \div 3 = 13$ (Factor of 3)
$13$ is already a prime number.
Step 3: Collect prime factors
Thus,
\[
156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13
\]
Step 4: Conclusion
Therefore, the prime factorisation of $156$ is $2^2 \times 3 \times 13$.
The correct answer is option (A).
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be:
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be: