Respuesta :
Part A:
For this case we have the following polynomial:
[tex]5x ^ 2a ^ 2 - 19xa ^ 2 - 4a ^ 2 [/tex]
First we make a common factor a^2:
We have then:
[tex]a ^ 2 (5x ^ 2 - 19x - 4) [/tex]
From here, we can factor the quadratic expression into parentheses.
We have then:
[tex]a ^ 2 ((5x + 1) (x-4)) [/tex]
Answer:
[tex]a ^ 2 ((5x + 1) (x-4)) [/tex]
Part B:
For this case we have the following polynomial:
[tex]x^2 + 14x + 49 [/tex]
We factor the expression.
To do this, we write two numbers that added are 14 and multiplied are 49.
We have then:
[tex](x + 7) (x + 7) [/tex]
Answer:
[tex](x + 7) (x + 7) [/tex]
Part C:
For this case we have the following polynomial:
[tex]x ^ 2 -100 [/tex]
We observe that we have a binomial, therefore, we must factor.
To do this, we write two numbers that added are 0 and multiplied are -100.
We have then:
[tex](x + 10) (x-10)[/tex]
Answer:
[tex](x + 10) (x-10)[/tex]
For this case we have the following polynomial:
[tex]5x ^ 2a ^ 2 - 19xa ^ 2 - 4a ^ 2 [/tex]
First we make a common factor a^2:
We have then:
[tex]a ^ 2 (5x ^ 2 - 19x - 4) [/tex]
From here, we can factor the quadratic expression into parentheses.
We have then:
[tex]a ^ 2 ((5x + 1) (x-4)) [/tex]
Answer:
[tex]a ^ 2 ((5x + 1) (x-4)) [/tex]
Part B:
For this case we have the following polynomial:
[tex]x^2 + 14x + 49 [/tex]
We factor the expression.
To do this, we write two numbers that added are 14 and multiplied are 49.
We have then:
[tex](x + 7) (x + 7) [/tex]
Answer:
[tex](x + 7) (x + 7) [/tex]
Part C:
For this case we have the following polynomial:
[tex]x ^ 2 -100 [/tex]
We observe that we have a binomial, therefore, we must factor.
To do this, we write two numbers that added are 0 and multiplied are -100.
We have then:
[tex](x + 10) (x-10)[/tex]
Answer:
[tex](x + 10) (x-10)[/tex]