Question
Factorise fully. 1.2.1. \( \quad\left(4 x^{6}-y^{4}\right) \)
Ask by Reese Craig. in South Africa
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The expression \( 4x^6 - y^4 \) can be fully factored as \( (2x^3 - y^2)(2x^3 + y^2) \).
Solution
We start with the expression
\[
4x^6 - y^4.
\]
**Step 1. Recognise the difference of squares.**
Notice that
\[
4x^6 = (2x^3)^2 \quad \text{and} \quad y^4 = (y^2)^2.
\]
Thus, we can write the expression as
\[
(2x^3)^2 - (y^2)^2.
\]
**Step 2. Apply the difference of squares factorisation formula.**
Recall the formula
\[
a^2 - b^2 = (a - b)(a + b).
\]
Let \( a = 2x^3 \) and \( b = y^2 \). Applying the formula gives
\[
(2x^3)^2 - (y^2)^2 = (2x^3 - y^2)(2x^3 + y^2).
\]
**Step 3. Conclude the factorisation.**
The fully factorised form of the expression is
\[
(2x^3 - y^2)(2x^3 + y^2).
\]
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Simplify this solution Mind Expander
To factorise \( 4x^6 - y^4 \), we first recognize it as a difference of squares. We can rewrite it as: \[ (2x^3)^2 - (y^2)^2 \] Using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \), we get: \[ (2x^3 - y^2)(2x^3 + y^2) \] Next, we check if either factor can be further factored. The factor \( 2x^3 - y^2 \) does not factor nicely with integers or simple rational numbers, and the same goes for \( 2x^3 + y^2 \). Therefore, the fully factorised form of \( 4x^6 - y^4 \) is: \[ (2x^3 - y^2)(2x^3 + y^2) \]
