# Determine the following and sketch: degree, LC, zeros(order 1,2,3), intervals positive and negative: f(x)=-4(x-3)(x+2)(2x-1); g(x)=0.5x(x+4)^2(2x-3)LC = Leading Coefficient Thank you!

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Notice that f(x) is a polynomial of third degree and you may find the leading coefficient opening the brackets such that:

`f(x) = -4(x-3)(x+2)(2x-1) => f(x) = (-4x + 12)(2x^2 - x + 4x - 2)`

`f(x) = (-4x + 12)(2x^2+3x-2) => f(x)= -8x^3 - 12x^2 + 8x + 24x^2 + 36x - 24`

`f(x) = -8x^3 + 12x^2 + 44x - 24`

**Notice that the coefficient of the term of highest degree is the leading coefficient and it is -8.**

You need to find the zeroes of the polynomial, hence, you need to solve the equation `-4(x-3)(x+2)(2x-1) = 0` such that:

`-4(x-3)(x+2)(2x-1) = 0`

`x - 3 = 0 => x_1 = 3`

`x + 2 = 0 => x_2 = -2`

`2x -1 = 0 => x_3 = 1/2`

Notice that if `x in [0,2]=> f(x)>0` and if `x in (-oo,0)U(3,oo) => f(x)<0` .

Notice that the highest degree of the given polynomial g(x) is 3.

`g(x)=0.5x(x+4)^2(2x-3)`

You need to find the leading coefficient, hence, you need to open the brackets such that:

` g(x)=0.5(x^2 + 8x + 16)(2x^2-3x)`

`g(x)=0.5(2x^4 - 3x^3 + 16x^3 - 24x^2 + 32x^2 - 48x)`

`g(x)=0.5(2x^4 + 13x^3 + 8x^2 - 48x)`

`g(x)= x^4 + 13/2 x^3 + 4x^2 - 24x`

Hence, evaluating the leading coefficient yields 1.

You need to find the zeroes of polynomial, hence, you need to solve `g(x)= 0` .

`0.5x(x+4)^2(2x-3) = 0`

`x_1 = 0`

`(x+4)^2 = 0 =>x_2 = x_3 = -4`

`2x - 3 = 0 => x_4 = 3/2`

Notice that if `x in(0,1], ` the function has negative values and if `x in (-oo,0)U(3/2,oo)` , the function has positive values.

f(x)=-4(x-3)(x+2)(2x-1)=-4(x-3)(2x^2+3x-2)=-4(2x^3+3x^2-2x-6x^2-9x+6)=-4(2x^3-3x^2-11x+6)= -8x^3+12x^2+44x-24

**Degree=Maximum power=3**

**Leading coefficient= -8**

In the interval positive

f(x)>o

-4(x-3)(x+2)(2x-1)>0

4(x-3)(x+2)(2x-1)<0

When x<-2;

4(x-3)(x+2)(2x-1)<0

When -2<x<1/2;

4(x-3)(x+2)(2x-1)>0

when 1/2<x<3;

4(x-3)(x+2)(2x-1)<0

When x>3;

4(x-3)(x+2)(2x-1)>0

**Therefore interval positive;**

**(x>3)U( -2<x<1/2)**

**and Interval negative;**

** (x<-2)U( 1/2<x<3)**

let g(x)=0.5x(x+4)^2(2x-3);

When you expand this,

**Maximum power=degree= 4**

**The LC = 0.5*1*2 = 1**

Let's consider the sign of g(x) with the given function;

When X<-4

0.5x(x+4)^2(2x-3)>0

When -4<x<0

0.5x(x+4)^2(2x-3)>0

When 0<x<3/2

0.5x(x+4)^2(2x-3)<0

When x>3/2

0.5x(x+4)^2(2x-3)>0

**Therefore interval positive;**

** (X<-4)U(-4<x<0) and x>2/3**

**Interval negative;**

**0<x<3/2**