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12. use patterns and structure the table shows the product when multipl…

Question

  1. use patterns and structure the table shows the product when multiplying two binomials. what is the relationship between the numbers in the factors and the terms in the product? binomials products (x + 3)(x + 4) $x^2 + 7x + 12$ (x + 2)(x - 5) $x^2 - 3x - 10$ (x - 3)(x - 5) $x^2 - 8x + 15$ 13. error analysis describe and correct the error a student made when multiplying two binomials. (2x + 2)(4x - 1) $8x^2 - 2$

Explanation:

Response
Question 12

Step1: Analyze the first binomial product

For \((x + 3)(x + 4)=x^{2}+7x + 12\), the coefficient of \(x\) in the product (\(7\)) is \(3 + 4\) (sum of the constant terms in the binomials), and the constant term (\(12\)) is \(3\times4\) (product of the constant terms).

Step2: Analyze the second binomial product

For \((x + 2)(x - 5)=x^{2}-3x - 10\), the coefficient of \(x\) (\(- 3\)) is \(2+( - 5)=2 - 5\), and the constant term (\(-10\)) is \(2\times(-5)\).

Step3: Analyze the third binomial product

For \((x - 3)(x - 5)=x^{2}-8x + 15\), the coefficient of \(x\) (\(-8\)) is \(-3+( - 5)=-3 - 5\), and the constant term (\(15\)) is \((-3)\times(-5)\).

Step1: Identify the error

The student forgot to use the distributive property (FOIL method) correctly. When multiplying \((2x + 2)(4x-1)\), the student only multiplied the first terms (\(2x\times4x = 8x^{2}\)) and the last terms (\(2\times(-1)=-2\)) but missed multiplying the outer and inner terms.

Step2: Correct the multiplication

Using the FOIL method:

  • First terms: \(2x\times4x = 8x^{2}\)
  • Outer terms: \(2x\times(-1)=-2x\)
  • Inner terms: \(2\times4x = 8x\)
  • Last terms: \(2\times(-1)=-2\)

Now, combine like - terms: \(8x^{2}-2x + 8x-2=8x^{2}+6x - 2\)

Answer:

When multiplying two binomials of the form \((x + a)(x + b)\) (where \(a\) and \(b\) can be positive or negative), the product is \(x^{2}+(a + b)x+ab\). That is, the coefficient of the \(x\) - term in the product is the sum of the constant terms in the binomial factors, and the constant term in the product is the product of the constant terms in the binomial factors.

Question 13