1 identify and factor the difference of two perfect squares.
2 look at the expression below.
$-4x^2 + 16x - 24 + 6x$
which of the following cannot be a pair of factors of the expression?
a. $-2, x - 3$
b. $-2, 2x - 3$
c. $2, 4 - x$
d. $2, 2x - 3$
3 michelle factored $x^2 - 6x + 9$ as follows:
$x^2 - 6x + 9 =$
$x^2 - 2(x \cdot 3) + 3^2 =$
$(x - 3)(x + 3)$
the correct factorization should be:
a. $(x - 3)(x + 3)$
b. $(x + 3)^2$
c. $(x - 3)^2$
4 which of these statements accurately describes the graph of the function $f(x) = x^2 + 20x + 100$ ?
a. the graph passes through the x-axis at -10.
b. the graph passes through the x-axis at 10.
c. the graph passes through the y-axis at -10.
d. the graph passes through the y-axis at 10.

Question

Accepted Answer

### Question 2
# Explanation:
## Step1: Simplify the expression
First, combine like terms in $-4x^2 + 16x - 24 + 6x$. The like terms for $x$ are $16x$ and $6x$, so adding them gives $22x$. The expression becomes $-4x^2 + 22x - 24$. We can factor out a $-2$ (or $2$) to simplify. Factoring out $-2$: $-2(2x^2 - 11x + 12)$? Wait, no, let's check again. Wait, original expression: $-4x^2 + 16x + 6x - 24 = -4x^2 + 22x - 24$. Alternatively, factor out $-2$: $-2(2x^2 - 11x + 12)$? Wait, maybe I made a mistake. Wait, let's factor out $-2$ from all terms: $-2(2x^2 - 11x + 12)$? No, wait, $-4x^2\div(-2)=2x^2$, $22x\div(-2)= -11x$, $-24\div(-2)=12$. Alternatively, factor out $2$: $2(-2x^2 + 11x - 12)$. Wait, maybe another approach. Let's factor the original expression. Let's first write it as $-4x^2 + 22x - 24$. Multiply numerator and denominator by $-1$ to make the leading coefficient positive: $4x^2 - 22x + 24$. Now, factor this quadratic. The discriminant is $22^2 - 4\times4\times24 = 484 - 384 = 100$. So roots are $\frac{22\pm10}{8}$, so $\frac{32}{8}=4$ and $\frac{12}{8}=\frac{3}{2}$. So $4x^2 -22x +24 = (4x - 6)(x - 4)$ (wait, no, using factoring: $4x^2 -22x +24 = 2(2x^2 -11x +12)$, and $2x^2 -11x +12$ factors as $(2x - 3)(x - 4)$? Wait, $(2x - 3)(x - 4)=2x^2 -8x -3x +12=2x^2 -11x +12$. Yes. So going back, the original expression $-4x^2 +22x -24 = -2(2x^2 -11x +12)= -2(2x - 3)(x - 4)$. Alternatively, we can write it as $2(-2x + 3)(x - 4)$ or $2(3 - 2x)(x - 4)$ or $2(2x - 3)(4 - x)$ (since $(x - 4)= - (4 - x)$, so $-2(2x - 3)(x - 4)=2(2x - 3)(4 - x)$). Now let's check each option:

- Option A: $-2$ and $x - 3$. Let's multiply $-2(x - 3)= -2x +6$. Does this divide into the original expression? Let's see, original expression is $-4x^2 +22x -24$. If we divide $-4x^2 +22x -24$ by $-2(x - 3)= -2x +6$, let's perform polynomial division. Dividend: $-4x^2 +22x -24$, divisor: $-2x +6$. Multiply divisor by $2x$: $-4x^2 +12x$. Subtract from dividend: $(-4x^2 +22x -24) - (-4x^2 +12x)=10x -24$. Now, divide $10x -24$ by $-2x +6$. Multiply divisor by $-5$: $10x -30$. Subtract: $(10x -24)-(10x -30)=6$. So there's a remainder, but wait, maybe I made a mistake. Wait, earlier factoring showed the expression is $2(2x - 3)(4 - x)$ or $-2(2x - 3)(x - 4)$. Let's check option B: $-2$ and $2x - 3$. Multiply $-2(2x - 3)= -4x +6$. Divide original expression $-4x^2 +22x -24$ by $-4x +6$. Dividend: $-4x^2 +22x -24$, divisor: $-4x +6$. Multiply divisor by $x$: $-4x^2 +6x$. Subtract: $(-4x^2 +22x -24)-(-4x^2 +6x)=16x -24$. Multiply divisor by $-4$: $16x -24$. Subtract: $0$. So it factors as $(-4x +6)(x - 4)= -2(2x - 3)(x - 4)$, which matches. Option C: $2$ and $4 - x$. Multiply $2(4 - x)=8 - 2x$. Divide original expression $-4x^2 +22x -24$ by $8 - 2x$ (which is $-2(x - 4)$). So dividing by $-2(x - 4)$ is same as multiplying by $-1/(2(x - 4))$. Wait, original expression is $-2(2x - 3)(x - 4)$, so $2(4 - x)=2(-(x - 4))= -2(x - 4)$. So $-2(2x - 3)(x - 4)= (2(4 - x))(2x - 3)$, since $-2(x - 4)=2(4 - x)$. So that works. Option D: $2$ and $2x - 3$. Multiply $2(2x - 3)=4x - 6$. Divide original expression $-4x^2 +22x -24$ by $4x - 6$. Dividend: $-4x^2 +22x -24$, divisor: $4x - 6$. Multiply divisor by $-x$: $-4x^2 +6x$. Subtract: $(-4x^2 +22x -24)-(-4x^2 +6x)=16x -24$. Now, divide $16x -24$ by $4x - 6$. Multiply divisor by $4$: $16x -24$. Subtract: $0$. Wait, but earlier we had the expression as $-2(2x - 3)(x - 4)$, and $2(2x - 3)=2(2x - 3)$, and $-(x - 4)=4 - x$. Wait, maybe I messed up the factoring. Wait, let's re-express the original expression: $-4x^2 +22x -24$. Let's factor out $-2$: $-2(2x^2 -11x +12)$. Now, $2x^2 -11x +12$ factors as $(2x - 3)(x - 4)$ (since $2x\times x=2x^2$, $2x\times(-4)=-8x$, $-3\times x=-3x$, $-3\times(-4)=12$; $-8x -3x=-11x$). So $-2(2x - 3)(x - 4)$. Now, let's check each option:

- Option A: $-2$ and $x - 3$. Multiply $-2(x - 3)= -2x +6$. If we multiply $-2(x - 3)$ with another factor, say $ax + b$, we get $-2ax^2 + (-2b + 6a)x -6b$. This should equal $-4x^2 +22x -24$. So $-2a = -4$ (so $a=2$), $-2b +6a=22$ (so $-2b +12=22$ → $-2b=10$ → $b=-5$), and $-6b=-24$ → $b=4$. But $b$ can't be both $-5$ and $4$, so this is not a factor pair. Wait, but earlier when we divided, we got a remainder, so A is not a factor pair? Wait, no, wait the options are pairs of factors. So the expression is $-4x^2 +22x -24$. Let's check each option by multiplying the pair:

- Option A: $-2 \times (x - 3) = -2x +6$. Now, if we multiply $-2x +6$ with another factor, say $2x - 4$ (since $-4x^2 +22x -24 = (-2x +6)(2x - 4)$? Let's check: $(-2x +6)(2x - 4)= -4x^2 +8x +12x -24= -4x^2 +20x -24$, which is not the original expression (original has $22x$). So that's wrong. Wait, I think I made a mistake in combining like terms. Original expression: $-4x^2 +16x -24 +6x$. $16x +6x=22x$, so $-4x^2 +22x -24$. Now, let's factor out $-2$: $-2(2x^2 -11x +12)$. Wait, $2x^2 -11x +12$ factors as $(2x - 3)(x - 4)$ (as $2x\times x=2x^2$, $2x\times(-4)=-8x$, $-3\times x=-3x$, $-3\times(-4)=12$; $-8x -3x=-11x$). So $-2(2x - 3)(x - 4)$. Now, let's check each option by multiplying the pair:

- Option A: $-2 \times (x - 3) = -2x +6$. Now, multiply $-2x +6$ with $2x - 4$: $-4x^2 +8x +12x -24 = -4x^2 +20x -24$ (not equal to original $-4x^2 +22x -24$). So A is not a factor pair? Wait, no, the original expression is $-4x^2 +22x -24$. Let's check Option B: $-2 \times (2x - 3) = -4x +6$. Multiply $-4x +6$ with $x - 4$: $-4x^2 +16x +6x -24 = -4x^2 +22x -24$ (which matches!). So B is a factor pair. Option C: $2 \times (4 - x) = 8 - 2x$. Multiply $8 - 2x$ with $2x - 3$: $16x -24 -4x^2 +6x = -4x^2 +22x -24$ (matches!). Option D: $2 \times (2x - 3) = 4x - 6$. Multiply $4x - 6$ with $-x + 4$ (since $-4x^2 +22x -24 = (4x - 6)(-x + 4)$): $4x\times(-x) +16x +6x -24 = -4x^2 +22x -24$ (matches!). So Option A is the one that cannot be a pair of factors.

## Step2: Verify each option
- Option A: Multiply $-2$ and $x - 3$ to get $-2x + 6$. Then, $(-2x + 6)(2x - 4) = -4x^2 + 8x + 12x - 24 = -4x^2 + 20x - 24$, which is not equal to the original expression $-4x^2 + 22x - 24$. So this pair does not multiply to the original expression.
- Option B: Multiply $-2$ and $2x - 3$ to get $-4x + 6$. Then, $(-4x + 6)(x - 4) = -4x^2 + 16x + 6x - 24 = -4x^2 + 22x - 24$, which matches.
- Option C: Multiply $2$ and $4 - x$ to get $8 - 2x$. Then, $(8 - 2x)(2x - 3) = 16x - 24 - 4x^2 + 6x = -4x^2 + 22x - 24$, which matches.
- Option D: Multiply $2$ and $2x - 3$ to get $4x - 6$. Then, $(4x - 6)(-x + 4) = -4x^2 + 16x + 6x - 24 = -4x^2 + 22x - 24$, which matches.

# Answer: A. $-2$, $x - 3$

### Question 3
# Explanation:
## Step1: Recall the perfect square trinomial formula
The perfect square trinomial formula is $a^2 - 2ab + b^2 = (a - b)^2$ and $a^2 + 2ab + b^2 = (a + b)^2$. For the expression $x^2 - 6x + 9$, we can identify $a = x$ and $b = 3$ because $x^2 = a^2$, $9 = b^2$, and $-6x = -2 \times x \times 3$ (which is $-2ab$).

## Step2: Apply the formula
Using the formula $a^2 - 2ab + b^2 = (a - b)^2$, substituting $a = x$ and $b = 3$, we get $(x - 3)^2$. Michelle incorrectly used the difference of squares formula $(a - b)(a + b) = a^2 - b^2$, which is for the difference of two squares, not a perfect square trinomial.

# Answer: C. $(x - 3)^2$

### Question 4
# Explanation:
## Step1: Factor the quadratic function
The function $f(x) = x^2 + 20x + 100$ is a perfect square trinomial. Using the formula $a^2 + 2ab + b^2 = (a + b)^2$, we identify $a = x$ and $b = 10$ (since $x^2 = a^2$, $100 = b^2$, and $20x = 2 \times x \times 10$ (which is $2ab$)). So $f(x) = (x + 10)^2$.

## Step2: Analyze the x-intercept and y-intercept
- **X-intercept**: The x-intercept occurs where $f(x) = 0$. So $(x + 10)^2 = 0$ implies $x + 10 = 0$ → $x = -10$. So the graph touches the x-axis at $x = -10$ (a repeated root).
- **Y-intercept**: The y-intercept occurs where $x = 0$. Substitute $x = 0$ into $f(x)$: $f(0) = 0^2 + 20(0) + 100 = 100$. So the graph passes through the y-axis at $100$, not $-10$ or $10$.

Now, check the options:
- Option A: The graph passes through the x-axis at $-10$. This is correct because the root is $x = -10$ (a repeated root, so it touches the x-axis here).
- Option B: The graph passes through the x-axis at $10$. Incorrect, since the root is $-10$.
- Option C: The graph passes through the y-axis at $-10$.

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QUESTION IMAGE

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Explanation:

Question 2

Step1: Simplify the expression

Step1: Recall the perfect square trinomial formula

Step2: Apply the formula

Step1: Factor the quadratic function

Step2: Analyze the x-intercept and y-intercept

Answer:

Question 3