Question

Questions 1-3 how do i factor those? Can you show the work and explain how?

Answer 1

1. To factor the first expression we have to take 3 as a common factor of the expression:

`3(n^2+3n+2)`

Now, we can factor the expression that is inside the parenthesis:

`3(n+2)(n+1)`

For this last step, we have to write two binomials, the first term of both will be n, the sign of the first binomial will be the first sign in the original expression (+). And the sign of the second binomial will be the product if the first and the second signs of the original expression (+*+=+)

The factored expression is 3(n+2)(n+1).

2. To factor this expression we have to rewrite it as follows:

`\begin{gathered} 28+x^2-11x \\ x^2-11x+28 \\ x^2-4x-7x+28 \end{gathered}`

Factor the two obtained binomials:

`\begin{gathered} (x^2-4x)+(-7x+28) \\ x(x-4)-7(x-4) \end{gathered}`

Take (x-4) as the common factor of the expression, to obtain the factored expression:

`(x-4)(x-7)`

The factored expression is (x-4)(x-7).

3. This expression can be factored using the factorization of perfect square trinomials. Find the square roots of the first and the third terms:

`9x^2-12x+4`

sqrt(9x^2)=3x.

sqrt(4)=2.

The factored expression is:

`(3x-2)^2`