algebra 2 assignment
factor the polynomial completely.
1. $12x^3 + 96x^2 + 192x = 0$

2. $108x^6 + 32x^3$

3. $x^3 + x^2 - 16x - 16$

4. $7x^{12} + 49x^9 + 70x^6$

5. the volume (in cubic feet) of a wooden crate is modeled by
$v = 2x^3 - 9x^2 + 10x$, where $x$ is the length (in feet). determine the values of $x$ for which the
model makes sense. explain your reasoning.

Question

algebra 2 assignment
factor the polynomial completely.
1. $12x^3 + 96x^2 + 192x = 0$

2. $108x^6 + 32x^3$

3. $x^3 + x^2 - 16x - 16$

4. $7x^{12} + 49x^9 + 70x^6$

5. the volume (in cubic feet) of a wooden crate is modeled by
$v = 2x^3 - 9x^2 + 10x$, where $x$ is the length (in feet). determine the values of $x$ for which the
model makes sense. explain your reasoning.

Accepted Answer

### Problem 1: $12x^3 + 96x^2 + 192x = 0$ # Explanation: ## Step 1: Factor out the GCF The greatest common factor (GCF) of $12x^3$, $96x^2$, and $192x$ is $12x$. Factoring it out: $12x(x^2 + 8x + 16) = 0$ ## Step 2: Factor the quadratic The quadratic $x^2 + 8x + 16$ is a perfect square trinomial, since $8x = 2 \cdot x \cdot 4$ and $16 = 4^2$. Thus, $x^2 + 8x + 16 = (x + 4)^2$. ## Step 3: Write the fully factored form Substituting back, the factored form is: $12x(x + 4)^2 = 0$ ### Problem 2: $108x^6 + 32x^3$ # Explanation: ## Step 1: Factor out the GCF The GCF of $108x^6$ and $32x^3$ is $4x^3$. Factoring it out: $4x^3(27x^3 + 8) = 0$ ## Step 2: Factor the sum of cubes The expression $27x^3 + 8$ is a sum of cubes, since $27x^3 = (3x)^3$ and $8 = 2^3$. Using the sum of cubes formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a = 3x$ and $b = 2$: $27x^3 + 8 = (3x + 2)(9x^2 - 6x + 4)$ ## Step 3: Write the fully factored form Substituting back, the factored form is: $4x^3(3x + 2)(9x^2 - 6x + 4)$ ### Problem 3: $x^3 + x^2 - 16x - 16$ # Explanation: ## Step 1: Group terms Group the first two and last two terms: $(x^3 + x^2) + (-16x - 16)$ ## Step 2: Factor out GCF from each group Factor $x^2$ from the first group and $-16$ from the second group: $x^2(x + 1) - 16(x + 1)$ ## Step 3: Factor out the common binomial Factor out $(x + 1)$ from both terms: $(x + 1)(x^2 - 16)$ ## Step 4: Factor the difference of squares The expression $x^2 - 16$ is a difference of squares ($x^2 - 4^2$), so factor it as $(x - 4)(x + 4)$: $(x + 1)(x - 4)(x + 4)$ ### Problem 4: $7x^{12} + 49x^9 + 70x^6$ # Explanation: ## Step 1: Factor out the GCF The GCF of $7x^{12}$, $49x^9$, and $70x^6$ is $7x^6$. Factoring it out: $7x^6(x^6 + 7x^3 + 10) = 0$ ## Step 2: Factor the quadratic in $x^3$ Let $y = x^3$, so the quadratic becomes $y^2 + 7y + 10$. Factor this: $y^2 + 7y + 10 = (y + 2)(y + 5)$ ## Step 3: Substitute back $y = x^3$ Substituting $y = x^3$ back, we get: $7x^6(x^3 + 2)(x^3 + 5)$ ### Problem 5: Volume $V = 2x^3 - 9x^2 + 10x$ # Explanation: ## Step 1: Factor the volume formula Factor out $x$ first: $V = x(2x^2 - 9x + 10)$ Then factor the quadratic $2x^2 - 9x + 10$. Find two numbers that multiply to $2 \cdot 10 = 20$ and add to $-9$ (these are $-4$ and $-5$): $2x^2 - 4x - 5x + 10 = 2x(x - 2) - 5(x - 2) = (2x - 5)(x - 2)$ Thus, $V = x(2x - 5)(x - 2)$. ## Step 2: Determine valid $x$ Since $x$ is a length, $x > 0$. Also, the other dimensions (from $2x - 5$ and $x - 2$) must be positive (as length cannot be negative). - For $x - 2 > 0$: $x > 2$ - For $2x - 5 > 0$: $x > \frac{5}{2} = 2.5$ The most restrictive condition is $x > 2.5$ (since $x > 2.5$ implies $x > 2$ and $x > 0$). However, we also check when factors could be zero (but length cannot be zero, so we exclude roots where volume would be zero or negative). The roots of $V = 0$ are $x = 0$, $x = 2$, and $x = \frac{5}{2}$. For $V > 0$ (valid volume), test intervals: - $0 < x < 2$: $V$ is positive (e.g., $x = 1$: $V = 2 - 9 + 10 = 3 > 0$) - $2 < x < 2.5$: $V$ is negative (e.g., $x = 2.2$: $V = 2(10.648) - 9(4.84) + 10(2.2) = 21.296 - 43.56 + 22 = -0.264 < 0$) - $x > 2.5$: $V$ is positive (e.g., $x = 3$: $V = 54 - 81 + 30 = 3 > 0$) But length must be positive, and the crate’s dimensions (from $x$, $2x - 5$, $x - 2$) must be positive (since length cannot be negative). Wait, actually, the dimensions are $x$, $2x - 5$, and $x - 2$. For all dimensions to be positive: $x > 0$, $2x - 5 > 0 \implies x > 2.5$, and $x - 2 > 0 \implies x > 2$. So the strictest is $x > 2.5$. However, when $0 < x < 2$, $2x - 5$ is negative (e.g., $x = 1$: $2(1) - 5 = -3$) and $x - 2$ is negative (e.g., $x = 1$: $1 - 2 = -1$), so the volume would be positive (negative $\times$ negative $\times$ positive = positive), but dimensions can’t be negative. Thus, the model makes sense when $x > 2.5$ (all dimensions positive) or $0 < x < 2$ (two dimensions negative, but volume positive—though physically, negative length doesn’t make sense, so we consider $x > 2.5$ as the valid domain for positive dimensions). # Answers: 1. $\boldsymbol{12x(x + 4)^2 = 0}$ 2. $\boldsymbol{4x^3(3x + 2)(9x^2 - 6x + 4)}$ 3. $\boldsymbol{(x + 1)(x - 4)(x + 4)}$ 4. $\boldsymbol{7x^6(x^3 + 2)(x^3 + 5)}$ 5. The model makes sense for $x > \frac{5}{2}$ (or $x > 2.5$) feet (since all dimensions must be positive) or $0 < x < 2$ feet (volume positive but dimensions negative, which is physically invalid, so the practical domain is $x > 2.5$ feet).

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: \(12x^3 + 96x^2 + 192x = 0\)

Step 1: Factor out the GCF

Step 2: Factor the quadratic

Step 3: Write the fully factored form

Problem 2: \(108x^6 + 32x^3\)

Step 1: Factor out the GCF

Step 2: Factor the sum of cubes

Step 3: Write the fully factored form

Problem 3: \(x^3 + x^2 - 16x - 16\)

Step 1: Group terms

Step 2: Factor out GCF from each group

Step 3: Factor out the common binomial

Step 4: Factor the difference of squares

Problem 4: \(7x^{12} + 49x^9 + 70x^6\)

Answer: