Year 12 Cambridge 3U Textbook Problems (1 Viewer)

[x.Exhaust.x]

Could anyone post their solutions to the following questions in their 3U textbook? Thanks.

Chapter 4: Polynomial Functions

Ex 4E Question 5c, 5d, 6a, 6b

 

[lyounamu]

Could anyone post their solutions to the following questions in their 3U textbook? Thanks.

Chapter 4: Polynomial Functions

Ex 4E Question 5c, 5d, 6a, 6b

Where? Which page? I don't think it is Ex 4E. Just realised that it is Year 12 textbook. I cannot help you then, sorry, (Don't have the textbook, >_<;;

 

[x.Exhaust.x]

Where? Which page? I don't think it is Ex 4E. Just realised that it is Year 12 textbook. I cannot help you then, sorry, (Don't have the textbook, >_<;;

Lol Namu. I'll post the questions up instead, making life easier :

1. Express x^3+2x^2-3x+1 as a polynomial in (x+1)

2. If the polynomials 2x^2+4x+4 and a(x+1)^2+b(x+2)^2+c(x+3)^2 are equal for three values of x, find a, b and c.

3. A polynomial of degree 3 has a double zero at 2. When x = 1 it takes the value 6 and when x = 3 it takes the value 8. Find the polynomial.

4. Two zeroes of a polynomial of degree 3 are 1 and -3. When x = 2 it takes the value -15 and when x = -1 it takes the value 36. Find the polynomial.

 

[donetha]

EX 4E

Q5.

c) (x+1)^3 - (x+1)^2 - 4(x+1) + 5

d) a= 3 b= -2 c= 1

Q6.

a) P(x)= (x-2)^2.(x+5)

b) P(x)= (x-1)(x+3)(2x-7)

Taken straight from the answers

 

[lyounamu]

Lol Namu. I'll post the questions up instead, making life easier :

1. Express x^3+2x^2-3x+1 as a polynomial in (x+1)

2. If the polynomials 2x^2+4x+4 and a(x+1)^2+b(x+2)^2+c(x+3)^2 are equal for three values of x, find a, b and c.

3. A polynomial of degree 3 has a double zero at 2. When x = 1 it takes the value 6 and when x = 3 it takes the value 8. Find the polynomial.

4. Two zeroes of a polynomial of degree 3 are 1 and -3. When x = 2 it takes the value -15 and when x = -1 it takes the value 36. Find the polynomial.

Thanks.

1. As tommykins said, only way to work this out is to substitute x+1 and manipulate your way. I got it in the end. It's not an easy question.

2. In this question, you substitute -1, -2 and -3 into the two equations separately. For example:

Let x = -1, then 2x^2 + 4x + 4 = 2 and a(x+1)^2 + b(x+2)^2 + c(x+3)^2 = b + 4c.

Then you will find that 2 = b + 4c because they are both equal.

Let x = -2, then a + c = 4 and let x = -3, then b + 4a = 10

So you have three equations:

1) b + 4c = 2
2) a + c = 4
3) b + 4a = 10

Manipulating equation 3), you get b = 10 - 4a ...(4)

Substitute this to the equation (1) and you get:
10 - 4a + 4c = 2
-4a + 4c = - 8
a - c = 2 ...(5)

Using the elimination method, compare this with equation (2). Then
you find the value of c which is 1. c = 1

Substitute the value of c to get a. a = 3.

Then the value of b can be found. b = - 2

3. P(x) = (x-2)^2 . (x-a)

P(1) = (1-2)^2 . (1-a)
= 1 - a which is equal to 6

1-a = 6, so a = -5

Therefore, P(x) = (x-2)^2 . (x+5)

 

[tommykins]

1. Express x^3+2x^2-3x+1 as a polynomial in (x+1)

Sub in (x+1) for all the x's, manipulate it so that your result is equal to x^3+2x^2-3x+1.

2. If the polynomials 2x^2+4x+4 and a(x+1)^2+b(x+2)^2+c(x+3)^2 are equal for three values of x, find a, b and c.

Expand a(x+1)^2+b(x+2)^2+c(x+3)^2 and equate the coefficients.

3. A polynomial of degree 3 has a double zero at 2. When x = 1 it takes the value 6 and when x = 3 it takes the value 8. Find the polynomial.

P(x) = (x-2)²(x-a) . We need to find a.

P(1) = (1-2)²(1-a) = (-1)²(1-a) = 1(1-a) = 1 - a = 6

Thus 1-a = 6, -a = 5, a = -5.

P(x) = (x-2)²(x+5)

4. Two zeroes of a polynomial of degree 3 are 1 and -3. When x = 2 it takes the value -15 and when x = -1 it takes the value 36. Find the polynomial.

P(x) = (x+3)(x-1)(x-a).

Sub in x = 2, put it in and solve it like the one above.

Spoiler

a=5

EDIT : Weird, subbing in x = -1, you get a completely different answer. a = 8 in that case.

Will recheck working.

 

[lolokay]

1. Express x^3+2x^2-3x+1 as a polynomial in (x+1)

never done this sort of stuff, so there might be a better way.

first divide the equation* by (x + 1)
= (x+1)(x^2 +x -4) + 5
so you have as the last two terms, -4(x+1) and +5
so you are left with x^2(x+1) + x(x+1)
= x^3 + 2x^2 + x
so you know that the coefficient for (x+1)^3 must be 1, which gives you
x^3 +3x^2 +3x +1, which is x^2 + 2x + 1 too many
so b(x+1)2 must equal -(x^2 + 2x + 1)
= -(x+1)^2

giving you (x+1)^3 - (x+1)^2 -4(x+1) +5

*you don't have to divide the equation at all. your method should be:
so you know that the coefficient for (x+1)^3 must be 1, which gives you
x^3 +3x^2 +3x +1, which is x^2 -6x too many
and then solve similarly for every lower power

 

[lolokay]

4. Two zeroes of a polynomial of degree 3 are 1 and -3. When x = 2 it takes the value -15 and when x = -1 it takes the value 36. Find the polynomial.

putting the polynomial as k(x-1)(x+3)(x+a)
when x = 2
5k(a + 2) = -15
k = -3/(a+2)

when x = -1
-4k(a - 1) = 36
-4*-3(a-1)/(a+2) = 36
a - 1 = 3(a + 2)
-2a = 7
a = -7/2

polynomial is therefore
2(x-1)(x+3)(x-7/2)
removing the fraction:
(x-1)(x+3)(2x-7)

Thanks.
2. In this question, you substitute -1, -2 and -3 into the two equations separately. For example:

how do you know to sub in those values? are you just taught that for these types of questions, that the equations will be equal when a(x+1) = 0, etc?

 

[midifile]

how do you know to sub in those values? are you just taught that for these types of questions, that the equations will be equal when a(x+1) = 0, etc?

You sub in -1, -2 and -3 so you can eliminate each of the letters and end up wilth simultaneous equations with only two unknowns in each. If you sub in other numbers, your simultaneous equations will have three unknowns in them making them harder to solve than if they only have two unknowns

But I prefer using tommykins method of equating the coefficients, although it is easier to make mistakes with equating coefficients if you are not good at expanding/havent learn binomial crap yet,

 

[lolokay]

You sub in -1, -2 and -3 so you can eliminate each of the letters and end up wilth simultaneous equations with only two unknowns in each. If you sub in other numbers, your simultaneous equations will have three unknowns in them making them harder to solve than if they only have two unknowns

But I prefer using tommykins method of equating the coefficients, although it is easier to make mistakes with equating coefficients if you are not good at expanding/havent learn binomial crap yet,

yeah I understand the equating coefficients thing, that was what I first thought to do. what threw me off was the question - it states the polynomials are equal for 3 values of x. Is this just another wasy of saying that the polynomials are equal for all x? since afaik you can't have 2 different quadratics equal at three points

 

[Aplus]

Where? Which page? I don't think it is Ex 4E. Just realised that it is Year 12 textbook. I cannot help you then, sorry, (Don't have the textbook, >_<;;

That library must be sick of you taking that book out

 

[lyounamu]

That library must be sick of you taking that book out

Nah, I am the only customer. So they love me.