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The problem is to expand the given binomial expressions. The expressions are: 1. $(x + 1)(x + 3)$

AlgebraPolynomial ExpansionBinomial ExpansionFOILDifference of Squares
2025/6/19

1. Problem Description

The problem is to expand the given binomial expressions. The expressions are:

1. $(x + 1)(x + 3)$

2. $(x + 3)(x + 2)$

3. $(x - 3)(x + 4)$

4. $(x + 5)(x - 2)$

5. $(a - 7)(a + 5)$

6. $(z + 9)(z - 2)$

7. $(k - 11)(k + 11)$

8. $(2x + 1)(x - 3)$

9. $(2y - 3)(y + 1)$

1

0. $(7y - 1)(7y + 1)$

2. Solution Steps

We use the distributive property (also known as FOIL) to expand the binomial expressions.
The general form is (a+b)(c+d)=a∗c+a∗d+b∗c+b∗d(a+b)(c+d) = a*c + a*d + b*c + b*d.

1. $(x + 1)(x + 3) = x*x + x*3 + 1*x + 1*3 = x^2 + 3x + x + 3 = x^2 + 4x + 3$

2. $(x + 3)(x + 2) = x*x + x*2 + 3*x + 3*2 = x^2 + 2x + 3x + 6 = x^2 + 5x + 6$

3. $(x - 3)(x + 4) = x*x + x*4 - 3*x - 3*4 = x^2 + 4x - 3x - 12 = x^2 + x - 12$

4. $(x + 5)(x - 2) = x*x + x*(-2) + 5*x + 5*(-2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10$

5. $(a - 7)(a + 5) = a*a + a*5 - 7*a - 7*5 = a^2 + 5a - 7a - 35 = a^2 - 2a - 35$

6. $(z + 9)(z - 2) = z*z + z*(-2) + 9*z + 9*(-2) = z^2 - 2z + 9z - 18 = z^2 + 7z - 18$

7. $(k - 11)(k + 11) = k*k + k*11 - 11*k - 11*11 = k^2 + 11k - 11k - 121 = k^2 - 121$. This is a difference of squares: $(a-b)(a+b) = a^2 - b^2$

8. $(2x + 1)(x - 3) = 2x*x + 2x*(-3) + 1*x + 1*(-3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$

9. $(2y - 3)(y + 1) = 2y*y + 2y*1 - 3*y - 3*1 = 2y^2 + 2y - 3y - 3 = 2y^2 - y - 3$

1

0. $(7y - 1)(7y + 1) = (7y)*(7y) + (7y)*(1) + (-1)*(7y) + (-1)*(1) = 49y^2 + 7y - 7y - 1 = 49y^2 - 1$. This is a difference of squares: $(a-b)(a+b) = a^2 - b^2$

3. Final Answer

1. $x^2 + 4x + 3$

2. $x^2 + 5x + 6$

3. $x^2 + x - 12$

4. $x^2 + 3x - 10$

5. $a^2 - 2a - 35$

6. $z^2 + 7z - 18$

7. $k^2 - 121$

8. $2x^2 - 5x - 3$

9. $2y^2 - y - 3$

1

0. $49y^2 - 1$

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