Question

4. (24 pts) For the following, if factorable, list the method(s) necessary to do so. If it is NOTfactorable, write prime and state how you determined that. Some will require morethan one method to factor completely. While it may be helpful to factor, you are notbeing asked for those answers; all credit is for simply stating the method(s) needed orthe detail that had you conclude prime.a) x² + 5x+7b) 4x² - 100c) 4x² - 11x +6d) x² - y²e) w* + 16f) 2xs-3x³+4x²-6

Answer 1

Part a

we have

x^2+5x+7

A quadratic equation with

a=1

b=5

c=7

so

b^2-4ac=5^2-4(1)(7)=25-28=-3

The given equation has no real roots

so is not factorable

Part b

4x^2-100

Remember that

a^2-b^2=(a+b)(a+b) -----> by difference of square

so

4x^2-100=(2x+10)(2x-10)

Part c

4x^2-11x+6

Solve the quadratic equation using the formula

a=4

b=-11

c=6

substitute in the formula

`x=(-(-11)\pm√(-11^2-4(4)(6)))/(2(4))`
`x=(11\pm5)/(8)`

The values of x are

x=2 and x=3/4

therefore

4x^2-11x+6=4(x-3/4)(x-2)=(4x-3)(x-2)

Part d

x^2-y^2

Applying the difference of squares

x^2-y^2=(x+y)(x-y)

Part e

w^2+16

w^2=-16

w=(+/-)4i

w=4i and w=-4i

w^2+16=(w-4i)(w+4i)

Part f

2x^5-3x^3+4x^2-6

using a graphing tool

The given equation has three real zeros

x=-1.23

x=-1.26

x=1.23

so

(x+1.23)(x+1.26)(x-1.23)=