Let's solve each problem step by step and explain the process.
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1. $(-x^3)(8x^5 + xy + 4y)$
We distribute $-x^3$ to each term inside the parentheses:
$$
= (-x^3)(8x^5) + (-x^3)(xy) + (-x^3)(4y)
$$
Now multiply:
- $(-x^3)(8x^5) = -8x^{3+5} = -8x^8$
- $(-x^3)(xy) = -x^{3+1}y = -x^4y$
- $(-x^3)(4y) = -4x^3y$
So, the result is:
$$
\boxed{-8x^8 - x^4y - 4x^3y}
$$
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2. $(x + 7)(x - 3)$
Use FOIL (First, Outer, Inner, Last):
- First: $x \cdot x = x^2$
- Outer: $x \cdot (-3) = -3x$
- Inner: $7 \cdot x = 7x$
- Last: $7 \cdot (-3) = -21$
Add them:
$$
x^2 - 3x + 7x - 21 = x^2 + 4x - 21
$$
$$
\boxed{x^2 + 4x - 21}
$$
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3. $(2x - 2)(4x + 1)$
Use FOIL:
- First: $2x \cdot 4x = 8x^2$
- Outer: $2x \cdot 1 = 2x$
- Inner: $-2 \cdot 4x = -8x$
- Last: $-2 \cdot 1 = -2$
Add:
$$
8x^2 + 2x - 8x - 2 = 8x^2 - 6x - 2
$$
$$
\boxed{8x^2 - 6x - 2}
$$
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4. $5x^3(2x^6 - xy^4 + 2y^6)$
Distribute $5x^3$ to each term:
- $5x^3 \cdot 2x^6 = 10x^{3+6} = 10x^9$
- $5x^3 \cdot (-xy^4) = -5x^{3+1}y^4 = -5x^4y^4$
- $5x^3 \cdot 2y^6 = 10x^3y^6$
$$
\boxed{10x^9 - 5x^4y^4 + 10x^3y^6}
$$
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5. $-7y^5(3y^2 + y + 2)$
Distribute $-7y^5$:
- $-7y^5 \cdot 3y^2 = -21y^{5+2} = -21y^7$
- $-7y^5 \cdot y = -7y^{5+1} = -7y^6$
- $-7y^5 \cdot 2 = -14y^5$
$$
\boxed{-21y^7 - 7y^6 - 14y^5}
$$
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6. $(3x - 4y)(6x + 8y)$
Use FOIL:
- First: $3x \cdot 6x = 18x^2$
- Outer: $3x \cdot 8y = 24xy$
- Inner: $-4y \cdot 6x = -24xy$
- Last: $-4y \cdot 8y = -32y^2$
Add:
$$
18x^2 + 24xy - 24xy - 32y^2 = 18x^2 - 32y^2
$$
$$
\boxed{18x^2 - 32y^2}
$$
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7. $(x + 10)^2$
This is a square of a binomial: $(a + b)^2 = a^2 + 2ab + b^2$
So:
$$
(x + 10)^2 = x^2 + 2(x)(10) + 10^2 = x^2 + 20x + 100
$$
$$
\boxed{x^2 + 20x + 100}
$$
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8. $(9x^2y^6)(x^3 - xy^4 + 2y^7)$
Distribute $9x^2y^6$ to each term:
- $9x^2y^6 \cdot x^3 = 9x^{2+3}y^6 = 9x^5y^6$
- $9x^2y^6 \cdot (-xy^4) = -9x^{2+1}y^{6+4} = -9x^3y^{10}$
- $9x^2y^6 \cdot 2y^7 = 18x^2y^{6+7} = 18x^2y^{13}$
$$
\boxed{9x^5y^6 - 9x^3y^{10} + 18x^2y^{13}}
$$
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9. Find the expression that represents the area of the triangle.
The triangle is right-angled, with legs:
- Height = $2x - 4$
- Base = $x + 8$
Area of a triangle:
$$
A = \frac{1}{2} \times \text{base} \times \text{height}
$$
So:
$$
A = \frac{1}{2}(x + 8)(2x - 4)
$$
Now multiply the binomials:
Use FOIL:
- $x \cdot 2x = 2x^2$
- $x \cdot (-4) = -4x$
- $8 \cdot 2x = 16x$
- $8 \cdot (-4) = -32$
Add:
$$
2x^2 - 4x + 16x - 32 = 2x^2 + 12x - 32
$$
Now multiply by $\frac{1}{2}$:
$$
\frac{1}{2}(2x^2 + 12x - 32) = x^2 + 6x - 16
$$
$$
\boxed{x^2 + 6x - 16}
$$
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✔ Final Answers:
1. $-8x^8 - x^4y - 4x^3y$
2. $x^2 + 4x - 21$
3. $8x^2 - 6x - 2$
4. $10x^9 - 5x^4y^4 + 10x^3y^6$
5. $-21y^7 - 7y^6 - 14y^5$
6. $18x^2 - 32y^2$
7. $x^2 + 20x + 100$
8. $9x^5y^6 - 9x^3y^{10} + 18x^2y^{13}$
9. $x^2 + 6x - 16$
Let me know if you'd like these written neatly on paper or formatted differently!
Parent Tip: Review the logic above to help your child master the concept of multiplying polynomials practice worksheet.