factor each completely when
13) $3x^2 + 19x + 6$

15) $7u^4 + 51u^2 + 54$

17) $2x^4 - 19x^2 + 45$

factor by grouping.
19) $x^3 - x^2 + x - 1 = 0$

21) $x^3 - 5x^2 - x + 5 = 0$

23) $x^4 + 4x^3 - x - 4$

Question

factor each completely when
13) $3x^2 + 19x + 6$

15) $7u^4 + 51u^2 + 54$

17) $2x^4 - 19x^2 + 45$

factor by grouping.
19) $x^3 - x^2 + x - 1 = 0$

21) $x^3 - 5x^2 - x + 5 = 0$

23) $x^4 + 4x^3 - x - 4$

Accepted Answer

### Problem 13: $3x^2 + 19x + 6$
# Explanation:
## Step 1: Find two numbers
Find two numbers that multiply to $3\times6 = 18$ and add up to $19$. The numbers are $18$ and $1$.
## Step 2: Split the middle term
Rewrite the middle term using these numbers: $3x^2+18x + x + 6$.
## Step 3: Group and factor
Group the first two and last two terms: $(3x^2 + 18x)+(x + 6)$. Factor out the GCF from each group: $3x(x + 6)+1(x + 6)$. Then factor out $(x + 6)$: $(3x + 1)(x + 6)$.
# Answer: $(3x + 1)(x + 6)$

### Problem 15: $7u^4 + 51u^2 + 54$
# Explanation:
## Step 1: Let $y = u^2$
Substitute $y = u^2$ to get $7y^2+51y + 54$.
## Step 2: Factor the quadratic
Find two numbers that multiply to $7\times54 = 378$ and add up to $51$. The numbers are $42$ and $9$. Rewrite the middle term: $7y^2+42y + 9y + 54$. Group: $(7y^2 + 42y)+(9y + 54)$. Factor: $7y(y + 6)+9(y + 6)=(7y + 9)(y + 6)$.
## Step 3: Substitute back $y = u^2$
We get $(7u^2 + 9)(u^2 + 6)$.
# Answer: $(7u^2 + 9)(u^2 + 6)$

### Problem 17: $2x^4 - 19x^2 + 45$
# Explanation:
## Step 1: Let $y = x^2$
Substitute $y = x^2$ to get $2y^2-19y + 45$.
## Step 2: Factor the quadratic
Find two numbers that multiply to $2\times45 = 90$ and add up to $-19$. The numbers are $-10$ and $-9$. Rewrite the middle term: $2y^2-10y-9y + 45$. Group: $(2y^2 - 10y)+(-9y + 45)$. Factor: $2y(y - 5)-9(y - 5)=(2y - 9)(y - 5)$.
## Step 3: Substitute back $y = x^2$
We get $(2x^2 - 9)(x^2 - 5)$. If we want to factor further over real numbers, we can write $2x^2 - 9=(\sqrt{2}x + 3)(\sqrt{2}x - 3)$ and $x^2 - 5=(x+\sqrt{5})(x - \sqrt{5})$, but the factored form with integer coefficients is $(2x^2 - 9)(x^2 - 5)$.
# Answer: $(2x^2 - 9)(x^2 - 5)$ (or $(\sqrt{2}x + 3)(\sqrt{2}x - 3)(x+\sqrt{5})(x - \sqrt{5})$ for real factors)

### Problem 19: $x^3 - x^2 + x - 1 = 0$ (Factor by grouping)
# Explanation:
## Step 1: Group terms
Group the first two and last two terms: $(x^3 - x^2)+(x - 1)$.
## Step 2: Factor each group
Factor out $x^2$ from the first group: $x^2(x - 1)+1(x - 1)$.
## Step 3: Factor out $(x - 1)$
Factor out $(x - 1)$ to get $(x^2 + 1)(x - 1)=0$.
# Answer: $(x^2 + 1)(x - 1)$

### Problem 21: $x^3 - 5x^2 - x + 5 = 0$ (Factor by grouping)
# Explanation:
## Step 1: Group terms
Group the first two and last two terms: $(x^3 - 5x^2)+(-x + 5)$.
## Step 2: Factor each group
Factor out $x^2$ from the first group and $-1$ from the second group: $x^2(x - 5)-1(x - 5)$.
## Step 3: Factor out $(x - 5)$
Factor out $(x - 5)$ to get $(x^2 - 1)(x - 5)$. Then factor $x^2 - 1$ as a difference of squares: $(x - 1)(x + 1)(x - 5)$.
# Answer: $(x - 1)(x + 1)(x - 5)$

### Problem 23: $x^4 + 4x^3 - x - 4$ (Factor by grouping)
# Explanation:
## Step 1: Group terms
Group the first two and last two terms: $(x^4 + 4x^3)+(-x - 4)$.
## Step 2: Factor each group
Factor out $x^3$ from the first group and $-1$ from the second group: $x^3(x + 4)-1(x + 4)$.
## Step 3: Factor out $(x + 4)$
Factor out $(x + 4)$ to get $(x^3 - 1)(x + 4)$. Then factor $x^3 - 1$ as a difference of cubes: $(x - 1)(x^2 + x + 1)(x + 4)$.
# Answer: $(x - 1)(x^2 + x + 1)(x + 4)$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 13: \(3x^2 + 19x + 6\)

Step 1: Find two numbers

Step 2: Split the middle term

Step 3: Group and factor

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

Step 2: Factor the quadratic

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

Step 1: Let \(y = x^2\)

Step 2: Factor the quadratic

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

Answer:

Problem 15: \(7u^4 + 51u^2 + 54\)