14. $x^4 - 12x^2 + 36$
15. $w^4 - 14w^2 - 32$
16. $k^3 + 7k^2 - 44k$
17. $2a^3 + 28a^2 + 96a$
18. $-x^3 + 4x^2 + 21x$
19. $m^6 - 7m^4 - 18m^2$
20. $9y^6 + 6y^4 + y^2$
21. $8c^4 + 10c^2 - 3$
22. $x^3 - 7x^2 + x - 7$
23. $4r^3 - 3r^2 - 4r + 3$
24. $3p^3 + 5p^2 - 12p - 20$
25. $15n^3 - 6n^2 - 25n + 10$

Question

Accepted Answer

Let's solve problem 14: $x^4 - 12x^2 + 36$

# Explanation:
## Step 1: Recognize as quadratic in $x^2$
Let $u = x^2$, then the expression becomes $u^2 - 12u + 36$.
## Step 2: Factor the quadratic
We need two numbers that multiply to $36$ and add to $-12$. Those numbers are $-6$ and $-6$. So, $u^2 - 12u + 36=(u - 6)^2$.
## Step 3: Substitute back $u = x^2$
$(x^2 - 6)^2$. We can also factor $x^2 - 6$ as a difference of squares: $x^2 - 6=(x - \sqrt{6})(x + \sqrt{6})$, so the fully factored form is $((x - \sqrt{6})(x + \sqrt{6}))^2=(x - \sqrt{6})^2(x + \sqrt{6})^2$, but if we consider integer coefficients only, $(x^2 - 6)^2$ is also acceptable.

# Answer:
$(x^2 - 6)^2$ (or $(x - \sqrt{6})^2(x + \sqrt{6})^2$)

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Problem 15: $w^4 - 14w^2 - 32$

# Explanation:
## Step 1: Let $u = w^2$
The expression becomes $u^2 - 14u - 32$.
## Step 2: Factor the quadratic
We need two numbers that multiply to $-32$ and add to $-14$. The numbers are $-16$ and $2$. So, $u^2 - 14u - 32=(u - 16)(u + 2)$.
## Step 3: Substitute back $u = w^2$
$(w^2 - 16)(w^2 + 2)$. Factor $w^2 - 16$ as a difference of squares: $w^2 - 16=(w - 4)(w + 4)$. So the fully factored form is $(w - 4)(w + 4)(w^2 + 2)$.

# Answer:
$(w - 4)(w + 4)(w^2 + 2)$

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Problem 16: $k^3 + 7k^2 - 44k$

# Explanation:
## Step 1: Factor out the GCF
The greatest common factor of $k^3$, $7k^2$, and $-44k$ is $k$. So, $k(k^2 + 7k - 44)$.
## Step 2: Factor the quadratic $k^2 + 7k - 44$
We need two numbers that multiply to $-44$ and add to $7$. The numbers are $11$ and $-4$. So, $k^2 + 7k - 44=(k + 11)(k - 4)$.

# Answer:
$k(k + 11)(k - 4)$

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Problem 17: $2a^3 + 28a^2 + 96a$

# Explanation:
## Step 1: Factor out the GCF
The GCF of $2a^3$, $28a^2$, and $96a$ is $2a$. So, $2a(a^2 + 14a + 48)$.
## Step 2: Factor the quadratic $a^2 + 14a + 48$
We need two numbers that multiply to $48$ and add to $14$. The numbers are $6$ and $8$. So, $a^2 + 14a + 48=(a + 6)(a + 8)$.

# Answer:
$2a(a + 6)(a + 8)$

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Problem 18: $-x^3 + 4x^2 + 21x$

# Explanation:
## Step 1: Factor out $-x$
$-x(x^2 - 4x - 21)$.
## Step 2: Factor the quadratic $x^2 - 4x - 21$
We need two numbers that multiply to $-21$ and add to $-4$. The numbers are $-7$ and $3$. So, $x^2 - 4x - 21=(x - 7)(x + 3)$.

# Answer:
$-x(x - 7)(x + 3)$

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Problem 19: $m^6 - 7m^4 - 18m^2$

# Explanation:
## Step 1: Factor out $m^2$
$m^2(m^4 - 7m^2 - 18)$.
## Step 2: Let $u = m^2$
The expression becomes $m^2(u^2 - 7u - 18)$.
## Step 3: Factor the quadratic $u^2 - 7u - 18$
We need two numbers that multiply to $-18$ and add to $-7$. The numbers are $-9$ and $2$. So, $u^2 - 7u - 18=(u - 9)(u + 2)$.
## Step 4: Substitute back $u = m^2$
$m^2(m^2 - 9)(m^2 + 2)$. Factor $m^2 - 9$ as a difference of squares: $m^2 - 9=(m - 3)(m + 3)$. So the fully factored form is $m^2(m - 3)(m + 3)(m^2 + 2)$.

# Answer:
$m^2(m - 3)(m + 3)(m^2 + 2)$

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Problem 20: $9y^6 + 6y^4 + y^2$

# Explanation:
## Step 1: Factor out $y^2$
$y^2(9y^4 + 6y^2 + 1)$.
## Step 2: Let $u = y^2$
The expression becomes $y^2(9u^2 + 6u + 1)$.
## Step 3: Factor the quadratic $9u^2 + 6u + 1$
This is a perfect square trinomial: $(3u + 1)^2$.
## Step 4: Substitute back $u = y^2$
$y^2(3y^2 + 1)^2$.

# Answer:
$y^2(3y^2 + 1)^2$

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Problem 21: $8c^4 + 10c^2 - 3$

# Explanation:
## Step 1: Let $u = c^2$
The expression becomes $8u^2 + 10u - 3$.
## Step 2: Factor the quadratic
We need two numbers that multiply to $8\times(-3)=-24$ and add to $10$. The numbers are $12$ and $-2$. Rewrite the middle term: $8u^2 + 12u - 2u - 3$.
## Step 3: Group and factor
$(8u^2 + 12u) + (-2u - 3)=4u(2u + 3)-1(2u + 3)=(4u - 1)(2u + 3)$.
## Step 4: Substitute back $u = c^2$
$(4c^2 - 1)(2c^2 + 3)$. Factor $4c^2 - 1$ as a difference of squares: $4c^2 - 1=(2c - 1)(2c + 1)$. So the fully factored form is $(2c - 1)(2c + 1)(2c^2 + 3)$.

# Answer:
$(2c - 1)(2c + 1)(2c^2 + 3)$

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Problem 22: $x^3 - 7x^2 + x - 7$

# Explanation:
## Step 1: Group terms
$(x^3 - 7x^2) + (x - 7)$.
## Step 2: Factor each group
$x^2(x - 7) + 1(x - 7)$.
## Step 3: Factor out $(x - 7)$
$(x - 7)(x^2 + 1)$.

# Answer:
$(x - 7)(x^2 + 1)$

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Problem 23: $4r^3 - 3r^2 - 4r + 3$

# Explanation:
## Step 1: Group terms
$(4r^3 - 3r^2) + (-4r + 3)$.
## Step 2: Factor each group
$r^2(4r - 3)-1(4r - 3)$.
## Step 3: Factor out $(4r - 3)$
$(4r - 3)(r^2 - 1)$.
## Step 4: Factor $r^2 - 1$ as a difference of squares
$(4r - 3)(r - 1)(r + 1)$.

# Answer:
$(4r - 3)(r - 1)(r + 1)$

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Problem 24: $3p^3 + 5p^2 - 12p - 20$

# Explanation:
## Step 1: Group terms
$(3p^3 + 5p^2) + (-12p - 20)$.
## Step 2: Factor each group
$p^2(3p + 5)-4(3p + 5)$.
## Step 3: Factor out $(3p + 5)$
$(3p + 5)(p^2 - 4)$.
## Step 4: Factor $p^2 - 4$ as a difference of squares
$(3p + 5)(p - 2)(p + 2)$.

# Answer:
$(3p + 5)(p - 2)(p + 2)$

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Problem 25: $15n^3 - 6n^2 - 25n + 10$

# Explanation:
## Step 1: Group terms
$(15n^3 - 6n^2) + (-25n + 10)$.
## Step 2: Factor each group
$3n^2(5n - 2)-5(5n - 2)$.
## Step 3: Factor out $(5n - 2)$
$(5n - 2)(3n^2 - 5)$.

# Answer:
$(5n - 2)(3n^2 - 5)$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Recognize as quadratic in \(x^2\)

Step 2: Factor the quadratic

Step 3: Substitute back \(u = x^2\)

Step 1: Let \(u = w^2\)

Step 2: Factor the quadratic

Step 3: Substitute back \(u = w^2\)

Step 1: Factor out the GCF

Step 2: Factor the quadratic \(k^2 + 7k - 44\)

Answer: