h) $(3x^4 + 5)^2$
j) $(3x^4 - 5)(3x^4 + 5)$
b) $\frac{20y^2 - 28y - 24}{5y + 3}$
d) $(x^4 + 2x^3 - 2x + 1) \div (1 - $
f) $(3y^3 + 11y^2 - 6y - 8) \div ($

Question

Accepted Answer

Let's solve each problem one by one. We'll start with part h, j, b, and then the division problems (though some division problems have incomplete denominators, we'll solve the ones with complete info first).

### Part h: Expand $\boldsymbol{(3x^4 + 5)^2}$
# Explanation:
## Step1: Recall the formula for squaring a binomial
The formula $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = 3x^4$ and $b = 5$.
## Step2: Apply the formula
First, calculate $a^2=(3x^4)^2 = 9x^8$. Then, $2ab = 2\times(3x^4)\times5 = 30x^4$. Finally, $b^2 = 5^2 = 25$.
Combine these terms: $(3x^4 + 5)^2=9x^8 + 30x^4 + 25$.
# Answer: $9x^8 + 30x^4 + 25$

### Part j: Multiply $\boldsymbol{(3x^4 - 5)(3x^4 + 5)}$
# Explanation:
## Step1: Recall the difference of squares formula
The formula $(a - b)(a + b)=a^2 - b^2$. Here, $a = 3x^4$ and $b = 5$.
## Step2: Apply the formula
Calculate $a^2=(3x^4)^2 = 9x^8$ and $b^2 = 5^2 = 25$. So, $(3x^4 - 5)(3x^4 + 5)=(3x^4)^2 - 5^2 = 9x^8 - 25$.
# Answer: $9x^8 - 25$

### Part b: Simplify $\boldsymbol{\frac{20y^2 - 28y - 24}{5y + 3}}$
# Explanation:
## Step1: Factor the numerator
First, factor out a common factor from the numerator. Let's factor 4 from $20y^2 - 28y - 24$: $4(5y^2 - 7y - 6)$. Now, factor the quadratic $5y^2 - 7y - 6$. We need two numbers that multiply to $5\times(-6)= -30$ and add to $-7$. Those numbers are $-10$ and $3$. So, $5y^2 - 10y + 3y - 6 = 5y(y - 2)+3(y - 2)=(5y + 3)(y - 2)$. Thus, the numerator becomes $4(5y + 3)(y - 2)$.
## Step2: Cancel common factors
The denominator is $5y + 3$. So, $\frac{4(5y + 3)(y - 2)}{5y + 3}=4(y - 2)=4y - 8$ (after canceling $5y + 3$ as long as $5y + 3
eq0$, i.e., $y
eq-\frac{3}{5}$).
# Answer: $4y - 8$ (for $y
eq-\frac{3}{5}$)

### Part d: Incomplete, but let's assume the denominator is $\boldsymbol{1 - x}$ (since it's a common divisor with the numerator's degree, though the original problem has a typo). Let's try polynomial long division for $\boldsymbol{(x^4 + 2x^3 - 2x + 1)\div(1 - x)}$
# Explanation:
First, rewrite the divisor as $-(x - 1)$, so the division becomes $-(x^4 + 2x^3 - 2x + 1)\div(x - 1)$. Now, use polynomial long division:
## Step1: Divide the leading term
Divide $x^4$ by $x$ to get $x^3$. Multiply $(x - 1)$ by $x^3$ to get $x^4 - x^3$. Subtract this from the numerator: $(x^4 + 2x^3 - 2x + 1)-(x^4 - x^3)=3x^3 - 2x + 1$.
## Step2: Next term
Divide $3x^3$ by $x$ to get $3x^2$. Multiply $(x - 1)$ by $3x^2$ to get $3x^3 - 3x^2$. Subtract: $(3x^3 - 2x + 1)-(3x^3 - 3x^2)=3x^2 - 2x + 1$.
## Step3: Next term
Divide $3x^2$ by $x$ to get $3x$. Multiply $(x - 1)$ by $3x$ to get $3x^2 - 3x$. Subtract: $(3x^2 - 2x + 1)-(3x^2 - 3x)=x + 1$.
## Step4: Next term
Divide $x$ by $x$ to get $1$. Multiply $(x - 1)$ by $1$ to get $x - 1$. Subtract: $(x + 1)-(x - 1)=2$.
So, the division $(x^4 + 2x^3 - 2x + 1)\div(1 - x)=-(x^3 + 3x^2 + 3x + 1) + \frac{2}{1 - x}$? Wait, no, let's do it properly. Wait, when we divide by $1 - x$, we can use synthetic division with root $x = 1$.
Using synthetic division for $x^4 + 2x^3 + 0x^2 - 2x + 1$ (we added $0x^2$ for clarity) and root $1$:
Coefficients: $1$ (x⁴), $2$ (x³), $0$ (x²), $-2$ (x), $1$ (constant)
Bring down the $1$.
Multiply by $1$: $1\times1 = 1$. Add to next coefficient: $2 + 1 = 3$.
Multiply by $1$: $3\times1 = 3$. Add to next coefficient: $0 + 3 = 3$.
Multiply by $1$: $3\times1 = 3$. Add to next coefficient: $-2 + 3 = 1$.
Multiply by $1$: $1\times1 = 1$. Add to last coefficient: $1 + 1 = 2$.
So, the quotient is $x^3 + 3x^2 + 3x + 1$ and the remainder is $2$. But since we divided by $1 - x$, the result is $-(x^3 + 3x^2 + 3x + 1) + \frac{2}{1 - x}$? Wait, no, synthetic division gives quotient $x^3 + 3x^2 + 3x + 1$ and remainder $2$ when dividing by $x - 1$, but we are dividing by $1 - x=-(x - 1)$, so the result is $-(x^3 + 3x^2 + 3x + 1)+\frac{2}{1 - x}$? This is getting complicated, and since the original problem's denominator is incomplete, we'll stop here.

### Part f: Incomplete, but let's assume the divisor is $\boldsymbol{y + 4}$ (a common factor, let's test $y = -4$ in the numerator: $3(-4)^3 + 11(-4)^2 - 6(-4) - 8 = 3(-64) + 11(16) + 24 - 8 = -192 + 176 + 16 = 0$, so $y + 4$ is a factor). Let's factor the numerator $3y^3 + 11y^2 - 6y - 8$ using $y = -4$ as a root.
# Explanation:
## Step1: Factor using root $y = -4$
Using synthetic division for $3y^3 + 11y^2 - 6y - 8$ with root $y = -4$:
Coefficients: $3$ (y³), $11$ (y²), $-6$ (y), $-8$ (constant)
Bring down the $3$.
Multiply by $-4$: $3\times(-4)= -12$. Add to next coefficient: $11 + (-12)= -1$.
Multiply by $-4$: $-1\times(-4)=4$. Add to next coefficient: $-6 + 4 = -2$.
Multiply by $-4$: $-2\times(-4)=8$. Add to last coefficient: $-8 + 8 = 0$. So, the numerator factors as $(y + 4)(3y^2 - y - 2)$. Now, factor $3y^2 - y - 2$: find two numbers that multiply to $3\times(-2)= -6$ and add to $-1$. Those numbers are $-3$ and $2$. So, $3y^2 - 3y + 2y - 2 = 3y(y - 1)+2(y - 1)=(3y + 2)(y - 1)$. Thus, the numerator is $(y + 4)(3y + 2)(y - 1)$. If the divisor is $y + 4$, then $\frac{(y + 4)(3y + 2)(y - 1)}{y + 4}=(3y + 2)(y - 1)=3y^2 - y - 2$ (for $y
eq -4$).
# Answer: If the divisor is $y + 4$, then $3y^2 - y - 2$ (for $y
eq -4$)

Since parts d and f have incomplete denominators, we solved the ones with complete information (h, j, b) and provided a possible solution for f assuming a common divisor.

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QUESTION IMAGE

Question

Explanation:

Part h: Expand \(\boldsymbol{(3x^4 + 5)^2}\)

Step1: Recall the formula for squaring a binomial

Step2: Apply the formula

Step1: Recall the difference of squares formula

Step2: Apply the formula

Step1: Factor the numerator

Step2: Cancel common factors

Answer:

Part j: Multiply \(\boldsymbol{(3x^4 - 5)(3x^4 + 5)}\)