(2) $$ x^{3}+2 x^{2}-3 x $$ (4) $$ 9 e^{2} f-12 e f^{2} $$ (6) $$ -15 y^{2}-3 y $$ (8) $$ 16 x^{4} y^{8}-8 x^{3} y^{7}-24 x^{2} y^{3} $$ (10) $$ \pi r^{2}+2 \pi r h $$ (12) $$ \frac{1}{4} a^{2}-\frac{1}{2} a b+\frac{3}{4} a^{2} b $$

Question

UpStudy AI · Accepted Answer

Sure! Let's factor each of the given algebraic expressions step by step.

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### (2) $$ x^{3} + 2x^{2} - 3x $$

**Step 1:** Factor out the greatest common factor (GCF), which is $$ x $$.

$$
x^{3} + 2x^{2} - 3x = x(x^{2} + 2x - 3)
$$

**Step 2:** Factor the quadratic expression $$ x^{2} + 2x - 3 $$.

We need two numbers that multiply to $$-3$$ and add to $$2$$. These numbers are $$3$$ and $$-1$$.

$$
x^{2} + 2x - 3 = (x + 3)(x - 1)
$$

**Final Factored Form:**

$$
x^{3} + 2x^{2} - 3x = x(x + 3)(x - 1)
$$

---

### (4) $$ 9e^{2}f - 12ef^{2} $$

**Step 1:** Factor out the GCF, which is $$ 3ef $$.

$$
9e^{2}f - 12ef^{2} = 3ef(3e - 4f)
$$

**Final Factored Form:**

$$
9e^{2}f - 12ef^{2} = 3ef(3e - 4f)
$$

---

### (6) $$ -15y^{2} - 3y $$

**Step 1:** Factor out the GCF, which is $$-3y$$.

$$
-15y^{2} - 3y = -3y(5y + 1)
$$

**Final Factored Form:**

$$
-15y^{2} - 3y = -3y(5y + 1)
$$

---

### (8) $$ 16x^{4}y^{8} - 8x^{3}y^{7} - 24x^{2}y^{3} $$

**Step 1:** Factor out the GCF, which is $$ 8x^{2}y^{3} $$.

$$
16x^{4}y^{8} - 8x^{3}y^{7} - 24x^{2}y^{3} = 8x^{2}y^{3}(2x^{2}y^{5} - x y^{4} - 3)
$$

**Step 2:** Check if the remaining trinomial can be factored further. In this case, it doesn't factor nicely, so we'll leave it as is.

**Final Factored Form:**

$$
16x^{4}y^{8} - 8x^{3}y^{7} - 24x^{2}y^{3} = 8x^{2}y^{3}(2x^{2}y^{5} - x y^{4} - 3)
$$

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### (10) $$ \pi r^{2} + 2\pi r h $$

**Step 1:** Factor out the GCF, which is $$ \pi r $$.

$$
\pi r^{2} + 2\pi r h = \pi r(r + 2h)
$$

**Final Factored Form:**

$$
\pi r^{2} + 2\pi r h = \pi r(r + 2h)
$$

---

### (12) $$ \frac{1}{4}a^{2} - \frac{1}{2}ab + \frac{3}{4}a^{2}b $$

**Step 1:** To make factoring easier, factor out the GCF. Notice that each term has at least $$ \frac{1}{4}a $$.

$$
\frac{1}{4}a^{2} - \frac{1}{2}ab + \frac{3}{4}a^{2}b = \frac{1}{4}a \left( a - 2b + 3a b \right)
$$

**Step 2:** Factor the expression inside the parentheses if possible. Here, we can rearrange the terms:

$$
a - 2b + 3a b = a(1 + 3b) - 2b
$$

This doesn't factor further neatly, so we'll leave it as is.

**Final Factored Form:**

$$
\frac{1}{4}a^{2} - \frac{1}{2}ab + \frac{3}{4}a^{2}b = \frac{1}{4}a(a - 2b + 3ab)
$$

---

If you have any specific questions about these factorizations or need further assistance with other expressions, feel free to ask!
Here are the factored forms of the given expressions: 1. $$ x^{3} + 2x^{2} - 3x = x(x + 3)(x - 1) $$ 2. $$ 9e^{2}f - 12ef^{2} = 3ef(3e - 4f) $$ 3. $$ -15y^{2} - 3y = -3y(5y + 1) $$ 4. $$ 16x^{4}y^{8} - 8x^{3}y^{7} - 24x^{2}y^{3} = 8x^{2}y^{3}(2x^{2}y^{5} - x y^{4} - 3) $$ 5. $$ \pi r^{2} + 2\pi r h = \pi r(r + 2h) $$ 6. $$ \frac{1}{4}a^{2} - \frac{1}{2}ab + \frac{3}{4}a^{2}b = \frac{1}{4}a(a - 2b + 3ab) $$

(2) \( x^{3}+2 x^{2}-3 x \) (4) \( 9 e^{2} f-12 e f^{2} \) (6) \( -15 y^{2}-3 y \) (8) \( 16 x^{4} y^{8}-8 x^{3} y^{7}-24 x^{2} y^{3} \) (10) \( \pi r^{2}+2 \pi r h \) (12) \( \frac{1}{4} a^{2}-\frac{1}{2} a b+\frac{3}{4} a^{2} b \)

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