Question
Factor completely.
y3–2y2+24y–48
Answers
Answered by
GPT-4o mini
To factor the polynomial \( y^3 - 2y^2 + 24y - 48 \) completely, we can use the method of grouping.
1. First, we can try to group the terms in pairs:
\[
(y^3 - 2y^2) + (24y - 48)
\]
2. Now, factor out the common factor from each group:
\[
y^2(y - 2) + 24(y - 2)
\]
3. Notice that \( (y - 2) \) is a common factor, so we can factor that out:
\[
(y - 2)(y^2 + 24)
\]
4. Now, the expression \( y^2 + 24 \) cannot be factored further over the real numbers because it does not have real roots. However, if you want to factor it over the complex numbers, you can express it as:
\[
y^2 + 24 = y^2 + (2\sqrt{6})^2 = (y - 2\sqrt{6}i)(y + 2\sqrt{6}i)
\]
5. Thus, the complete factorization over the complex numbers is:
\[
(y - 2)(y - 2\sqrt{6}i)(y + 2\sqrt{6}i)
\]
However, over the real numbers, the completely factored form of \( y^3 - 2y^2 + 24y - 48 \) is:
\[
(y - 2)(y^2 + 24)
\]
1. First, we can try to group the terms in pairs:
\[
(y^3 - 2y^2) + (24y - 48)
\]
2. Now, factor out the common factor from each group:
\[
y^2(y - 2) + 24(y - 2)
\]
3. Notice that \( (y - 2) \) is a common factor, so we can factor that out:
\[
(y - 2)(y^2 + 24)
\]
4. Now, the expression \( y^2 + 24 \) cannot be factored further over the real numbers because it does not have real roots. However, if you want to factor it over the complex numbers, you can express it as:
\[
y^2 + 24 = y^2 + (2\sqrt{6})^2 = (y - 2\sqrt{6}i)(y + 2\sqrt{6}i)
\]
5. Thus, the complete factorization over the complex numbers is:
\[
(y - 2)(y - 2\sqrt{6}i)(y + 2\sqrt{6}i)
\]
However, over the real numbers, the completely factored form of \( y^3 - 2y^2 + 24y - 48 \) is:
\[
(y - 2)(y^2 + 24)
\]