# polynomial question plz answer the last question part 2 and 3 the equation and the answer  http://img26.imageshack.us/i/dsc04434l.jpg/ plz hurry i have the homework 2morrow plz thank u all

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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1) For finding the coefficients a and b, the given polynomial has to be equal to the polynomial x^3 + 2x^2 - x - 2, for that we have to put in a equality relation the corresponding coefficients.

Let's see how:

x^3 + (a-1)x^2 - x + b+1 = x^3 + 2x^2 - x - 2

The coefficient of x^2, from the left side polynomial, is (a-1) and it has to be equal to the corresponding coefficient of x^2, from the right side, which is 2.

a-1=2

a-1+1=2+1

a=3

The same we'll proceed with the terms:

b+1=-2

b+1-1=-2-1

b=-3

2) For finding the coefficients a and b, we have to follow the constraint, which states that the polynomial p(x) is divisible by x and x+2.

If p(x) is divisible by x, that means that if p(x) will be divided by x, the reminder will be 0.

The constraint is mathematically written:

p(0)=0

For calculating p(0), we'll substitute x from polynomial by 0.

p(0) = 0^3 + (a-1)0^2 - 0 + b+1

b+1=0

b=-1

If p(x) is divisible by x+2, that means that if p(x) will be divided by x+2, the reminder will be 0.

We'll write the division with reminder:

p(x)=(x+2)*Q(x) + R(x)

We notice that we substitute x with -2, we'll have:

p(-2)=(-2+2)*Q(-2) + R(-2), where R(-2)=0, because p(x) is divisible by x+2.

p(-2)=0*Q(-2)+0

p(-2) = (-2)^3 + (a-1)(-2)^2 - (-2) -1+1

p(-2) = -8 + 4a - 4 + 2

But p(-2)=0

4a-10=0

4a=10

a=10/4

a=5/2

3) If x+1 is a factor of p(x), that means that p(x) will be divided by x+1 and the reminder will be 0.

We'll write the division with reminder:

p(x)=(x+1)*Q(x) + R(x)

We notice that we substitute x with -1, we'll have:

p(-1)=(-1+1)*Q(-1) + R(-1), where R(-1)=0, because p(x) is divisible by x+1.

p(-1) = (-1)^3 + (a-1)(-1)^2 - (-1) +b+1

p(-1) = -1+a-1+1+b+1

But p(-1) = 0, so a+b=0

If p(x) is divided by x+3 and the reminder is 30, we'll write:

p(x)=(x+3)*Q(x) + R(x)

We notice that we substitute x with -3, we'll have:

p(-3)=(-3+3)*Q(-3) + R(-3), where R(-3)=30.

p(-3) = (-3)^3 + (a-1)(-3)^2 - (-3) +b+1

But p(-3)=30

-27+9a-9+3+b+1=30, but a=-b

-32+9a-a=30

8a=30+32

8a=62

a=62/8

a=31/4 and b=-31/4

neela | High School Teacher | (Level 3) Valedictorian

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2)

a) To solve   |x+5|-12 =3 .

Solution:

|x+5| = 12+3 =15.

If x+5 < 0 then x<-5 for whick |x+5| = -x-5 =15 or -5-15 = x. Or x= -20.

If x+5 > = 0, then |x+5| = x+5 -12 = 3 . Or x= 10

So x= 10 or x = -20

b)

5^(2x+1) = 625 = 5^4. Comparing the powers as the base is equal , 2x+1 = 4 . Or 2x= 4-1=3. So x = 3/2 =1.5.

c)

3^(2x+3).9^x = 81 . Or

3^(2x)* 9^x = 81/27 = 3. Or

3^(2x)*3^(2x) = 3. Or

3^(4x) = 3^1. Comparing the indices, 4x=1. Or x = 1/4

d)|3x+7| > 4.

If 3x+7 < 7, Then |3x+7| > 4 implies  for x<-7/3, -3x-7  > 4 Or -3x > 4+7 = 11. Or

3x<-11. Or

x <-11/3

Also if |3x+7| > 0 , then for x > -7/3,  |3x+7| > 4 implies

3x+7 > 4. Or 3x > 4-7 = -3. Or 3x > -3. Or x > -1.

So the solution set x <-11/3 Or x > -1. Or  x should not belong to the closed interval [ -7/3 ,-1] and x is out side this interval.

e (x^2-4)/(x+3) > 0. Or

(x+2)(x-2)/(x+3) > 0.

Lety f(x) = (x+2)(x-2)/(x+3) > 0 for x >2

For  x=2, f(x) = 0.

For -2<x<2 , f(x) = (+)(-)/(+) is -ve so -2<x<2 is not a solution.

When x = -2 f(x) = 0.

For  -3<x< -2, f(x) = (-)(-)/(+)  is positive. So -3<x<-2 is a solution.

For x=-3 , denominator is 0 f(x) is undefined.

For x < -3, f(x) = (-)(-)/(-) is negative.

Combining, -3<x<-2  Or x>2 is the solution.

3

x+2y-4 < 0 and x+2y + 2  > 0 . We can rewrite these equations in the slope intercept form as below:

y < (-1/2)x+2..........(1) and

y > (-1/2)x-1...........(2)

The inequality holds good for any point in between these two parallel lines.

modioh | Student, Grade 11 | (Level 1) eNoter

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