Question

I need help with this ​

Answer 1

1) The factorized form of the polynomial is
`(x-5)\cdot (x+2) = 0`.

2) The factorized form of the polynomial is
`(x+6)\cdot (x-4) = 0`.

3)
`r_(1) = -3`,
`r_(2) = -2`. (Option D)

4)
`r_(1) = -7`,
`r_(2) = 5`. (Option A)

Explanation:

All exercise are case of factorization of second grade polynomials of the form
`x^(2) +(-r_(1) - r_(2))\cdot x + r_(1)\cdot r_(2)`, where
`r_(1)` and
`r_(2)` are the two roots of the polynomial. Now we proceed to solve each polynomial:

1)
`x^(2)-3\cdot x-10 = 0`

In this case, the coefficients have the following characteristics:

`-r_(1)-r_(2) = -3`

`r_(1)\cdot r_(2) = -10`

The solution of this system of nonlinear equations is:
`r_(1) = 5`,
`r_(2) = -2`.

Then, the factorized form of the polynomial is:

`(x-5)\cdot (x+2) = 0`

2)
`x^(2)+2\cdot x -24 = 0`

In this case, the coefficients have the following characteristics:

`-r_(1)-r_(2) = 2`

`r_(1)\cdot r_(2) = -24`

The solution of this system of nonlinear equations is:
`r_(1) = -6`,
`r_(2) = 4`.

Then, the factorized form of the polynomial is:

`(x+6)\cdot (x-4) = 0`

3)
`x^(2) + 5\cdot x +6 = 0`

In this case, the coefficients have the following characteristics:

`-r_(1)-r_(2) = 5`

`r_(1)\cdot r_(2) = 6`

The solution of this system of nonlinear equations is:
`r_(1) = -3`,
`r_(2) = -2`.

Hence, the correct answer is D.

4)
`x^(2)+2\cdot x -35 = 0`

In this case, the coefficients have the following characteristics:

`-r_(1)-r_(2) = 2`

`r_(1)\cdot r_(2) = -35`

The solution of this system of nonlinear equations is:
`r_(1) = -7`,
`r_(2) = 5`.

Hence, the correct answer is A.