**Factor 4y**^{2} - 36y^{6}
There is a **common factor** of 4y^{2}
that can be factored out first in this problem, to make the
problem easier.
**4y**^{2} (1 - 9y^{4})
In the factor (1 - 9y^{4}), 1 and 9y^{4} are
perfect squares (their coefficients are perfect squares and
their exponents are even numbers).
Since subtraction is occurring between these squares, this
expression is the difference
of two squares.
What times itself will give 1? The answer is
1.
What times itself will give 9y^{4} ? The answer
is 3y^{2} .
The factors are (1 + 3y^{2})
and (1 - 3y^{2}).
**Answer: 4y**^{2} (1 +
3y^{2})
(1 - 3y^{2})
or 4y^{2} (1 -
3y^{2})
(1 + 3y^{2})
** ****__________________________________________________________**
If you did not see the **common
factor,** you can begin with observing the perfect squares. Both 4y^{2} and 36y^{6}
are perfect squares (their coefficients are perfect squares and
their exponents are even numbers). Since subtraction is
occurring between these squares, this expression is the difference
of two squares.
What times itself will give 4y^{2}
? The answer is 2y.
What times itself will give 36y^{6} ? The answer
is 6y^{3} .
The factors are (2y + 6y^{3})
and (2y - 6y^{3}).
Answer: (2y + 6y^{3})
(2y - 6y^{3})
or (2y - 6y^{3})
(2y + 6y^{3})
These answers can be further factored as each contains a common
factor of 2y:
2y (1 + 3y^{2}) • 2y
(1 - 3y^{2}) = 4y^{2} (1 + 3y^{2})
(1 - 3y^{2}) |