Thread: algebra polynomial help

Basically I've got nine questions, mixed subjects, from Heinemann Pure Maths 3, Review Exercise 1, and if anyone can do ANY of them it would be very helpful and I'd be so grateful.

1) f(x)= x^3+ax^2+bx+6
Find, in terms of a and b, the remainder when f(x) is divided by x-2 and x+3. Given that these remainders are equal, express a in terms of b.

2) Expand (1-4x)^1/4 in ascending powers of x up to and including the term x^3, simplifying each coefficient.

3) Find the coordinates of the centre and radius of the circle whose equation is x^2+y^2-16x-12y+96=0
Also find the least and greatest distances of the origin O from the circumference of the circle.

4)a) Given that (x+1) is a factor of the expression (2x^3+ax^2-5x-2), find the value of the constant a. Show that, with this value of a, (x-2) is another factor of this expression and hence, or otherwise, factorise the expression completely.
b) When divided by (x-2) the expression (x^3+x^2+2x+2) leaves a remainder R. Find the value of R.

5) The population, p, of insects on an island, t hours after midday, is given by p=1000e^(kt) where k is a constant. Given that when t=0, the rate of change of the population with respect to time is 100 per hour,
a) Find k
b) Find the population when t=6

6)Differentiate with respect to x:
a) (sinx)/e^x
b) ln(1+(tan^2)x)

7) When a metal cube is heated, the length of each edge increases at the rate of 0.03cm/s. Find the rate of increase, in cm^2/s, of the total surface area of the cube, when the length of each edge is 8cm.

8) Given that (x-2) is a factor of f(x) where f(x)=x^3-x^2+Ax+B
find an equation satisfied by the constants A and B.
Given that when f(x) is divided by (x-3) the remainder is 10, find the second equation satisfied by A and B.
Solve your equations to find A and B.
Using your values for A and B, find 3 values of x for which f(x)=0

9) Find the coordinates of the turning points on the curve with the equation
y^3+3xy^2-x^3=3


Thank you in advance

Hello, Miranda!

#9 is a messy one . . .

9) Find the coordinates of the turning points on the curve with the equation:
. . . y³ + 3xy² - x³ .= .3

Differentiate implicitly: .3y²y' + 6xyy' + 3y² - 3x² .= .0

Then we have: .3y²y' + 6xyy' .= .3y² - 3x²

Factor: .3y(y + 2x)y' .= .3(y² - x²)

. . . . . . . . . . . .y² - x²
Hence: . y' .= .-----------
. . . . . . . . . . .y(y + 2x)

The derivative equals zero when its numerator equals zero:
. . y² - x² .= .0 . → . y = ±x


y = x: .Substitute into the original equation.
. . x³ + 3x·x² - x³ .= .3 . → . x³ = 1 . → . x = 1 . → . y = 1

One turning point is: .(1, 1)


y = -x: .Substitute into the original equation.
. . -x³ + 3x·x² - x³ .= .3 . → . x³ = 3 . → . x = cbrt{3} . → . y = -cbrt{3}

Another turning point is: .(cbrt{3}, -cbrt{3})

Originally Posted by Miranda

...

3) Find the coordinates of the centre and radius of the circle whose equation is x^2+y^2-16x-12y+96=0
Also find the least and greatest distances of the origin O from the circumference of the circle.
...

Hello, Miranda,

rearrange the equation and complete the squares:

x² + y² - 16x - 12y= -96
x² - 16x + 64+ y²- 12y+ 36= -96 + 64 + 36
(x - 8)² + (y - 6)² = 2². Thus the center is C(8, 6) and the radius is 2.

The distance origin center is 10. (Use Pythagoran theorem)
Therefore the nearest point of the circle to the origin must be 2 units from C, that is 1/5 of the total distance OC.
Use similar right triangles and you'll get N(32/5, 24,5), that means N lies at 4/5 of the total distance. The most distant point of the circle to the origin must be 2 units from C, that means D is at 6/5 of the total distance: D(48/5, 36/5)

EB

Originally Posted by Miranda

...

5) The population, p, of insects on an island, t hours after midday, is given by p=1000e^(kt) where k is a constant. Given that when t=0, the rate of change of the population with respect to time is 100 per hour,
a) Find k
b) Find the population when t=6

...

Hello, Miranda,

the rate of change is calculated by the first derivative:

p(t) = 1000*e^(k*t) ===> p'(t) = 1000*k*e^(k*t) (chain rule)

We know that
p'(0) = 1000*k = 100 ===> k = 0.1

Therefore:
p(t) = 1000*e^(0.1*t)
Now plug in t = 6 and calculate p(6):

p(6) = 1000*e^(0.6) ≈ 1822

EB

Originally Posted by Miranda

...
1) f(x)= x^3+ax^2+bx+6
Find, in terms of a and b, the remainder when f(x) is divided by x-2 and x+3. Given that these remainders are equal, express a in terms of b.
...

Hello, Miranda,

you have to do long division twice:

Code:

(x³ + ax² + bx + 6) ÷ (x - 2) = x²+(a+2)x+2a+b+4
 x³ - 2x²
--------
      (a+2)x² + bx
      (a+2)x²-2(a+2)x
      ---------------
                  (2a+b+4)x + 6
                  (2a+b+4)x-2(2a+b+4)
                  ---------------------
                               4a+2b+14

The remainder is 4a + 2b + 14

Code:

(x³ + ax² + bx + 6) ÷ (x + 3) = x²+(a-3)x+b-3a+9
 x³ + 3x²
--------
      (a-3)x² + bx
      (a-3)x²+3(a-3)x
      ---------------
                  (b-3a-9)x + 6
                  (b-3a+9)x+3b-9a+27
                  ---------------------
                               9a-3b-21

The remainder is 9a - 3b - 21

According to your problem these remainders are equal:

9a - 3b - 21 = 4a + 2b + 14
5a = 5b +35
a = b+7

EB

Hello, Miranda!

Here's #8 . . .
You're expected to be familiar with the Factor Theorem and Remainder Theorem.

8) Given that (x - 2) is a factor of: .f(x) .= .x³ - x² + Ax + B

(a) Find an equation satisfied by the constants A and B.

(b) Given that when f(x) is divided by (x-3) the remainder is 10,
. . .find a second equation satisfied by A and B.

(c) Solve your equations to find A and B.

(d) Using your values for A and B, find 3 values of x for which f(x) = 0.

(a) If (x - 2) is a factor of f(x), then f(2) = 0.
. . .2³ - 2² + 2A + B .= .0 . → . 2A + B .= .-4 .[1]

(b) If f(x) ÷ (x - 3) has a remainder of 10, then f(3) = 10
. . .3³ - 3² + 3A + B .= .10 . → . 3A + B .= .-8 .[2]

(c) Subtract [1] from [2]: .A = -4
. . .Substitute into [1]: .2(-4) + B .= .-4 . → . B = 4

(d) We are told that (x - 2) is a factor of f(x) .= .x³ - x² - 4x + 4
. . .Then: .f(x) .= .(x - 2)(x² + x - 2) .= .(x - 2)(x - 1)(x + 2)
. . .The three values are: 2, 1, -2

Hello, Miranda!

If you know the Remainder Theorem, you don't need long division for #1.

1) .f(x) .= .x³ + ax² + bx + 6

Find, in terms of a and b, the remainder when f(x) is divided by x-2 and x+3.

Given that these remainders are equal, express a in terms of b.

If f(x) is divided by x -2, the remainder is f(2).

. . f(2) .= .2³ + a·2² + b·2 + 6 .= .4a + 2b + 14


If f(x) is divided by x + 3, the remainder is f(-3).

. . f(-3) .= .(-3)³ + a(-3)² + b(-3) + 6 .= .9a - 3b - 21


If these remainders are equal: .4a + 2b + 14 .= .9a - 3b - 21 . → . a .= .b + 7