Example 2: The formula for finding volume is V = (pi)r^2h
(a) Make h the subject of formula.
(b) Make r the subject of formula

(a) V = (pi)r²h
Divide both sides by (pi)r²
H = V / (pi)r2

b. V = (pi)r²h
Divide both sides by (pi)h
R² = V / (pi)h
Take the square root of both sides
R = Sqrt.{V / (pi)h}

CLASSWORK:
2. Make x the subject of formula if y = 4ax²

Example 3: Make t the subject of formula in s = ur + 1/2tr²

s = ur + 1/2tr²
Multiply both sides by 2
2s = 2ur + tr²
Subtract 2ur from both sides
2s – 2ur = tr²
Divide both sides by t
(2s – 2ur) / t = r²
Find the square root of both sides
R = sqrt{(2s – 2ur) / t}

CLASSWORK
3. Make n the subject of formula if m = 8a + mn³

Example 4. Given that V² – gh = g{sqrt(c² + h²), make h the subject of formula

V² – gh = g{sqrt(c² + h²)}
Divide both sides by g
V² / g – h = sqrt(c² + h²)
Square both sides
(V² / g – h)² = c² + h²
Subtract c² from both sides

H² = (V² / g – h)² – c²
Find the square root of both sides
H = sqrt{(V² / g – h)² – c²}
H = V²/g – h – c

CLASSWORK
4. Given r = 2(pi)sqrt(1/t) make t the subject of formula

THE END

thankyouJesus:
This is a direct copy, I need to be sure people still visit this nairaland for knowledge to avoid the math gurus wasting their time. Just 2 or more comments will do.
www.nairaland.com/modifypost?redirect=%2Fthankyoujesus%2Fposts%2F16%2331931741&post=31931741

i am following. pls we need physics gurus also.

Geofavor:

i am following. pls we need physics gurus also.

Okay, lets fire on. Could you help with a direct link to physics thread?

Mathemagician1:
Example 2: The formula for finding volume is V = (pi)r^2h
(a) Make h the subject of formula.
(b) Make r the subject of formula

(a) V = (pi)r^2h
Divide both sides by (pi)r^2
H = V / (pi)r^2

b. V = (pi)r^2h
Divide both sides by (pi)h
R^2 = V / (pi)h
Take the square root of both sides
R = Sqrt.{V / (pi)h}

CLASSWORK:
2. Make x the subject of formula if y = 4ax^2

Example 3: Make t the subject of formula in s = ur + 1/2tr^2

s = ur + 1/2tr^2
Multiply both sides by 2
2s = 2ur + tr^2
Subtract 2ur from both sides
2s – 2ur = tr^2
Divide both sides by t
(2s – 2ur) / t = r^2
Find the square root of both sides
R = sqrt{(2s – 2ur) / 2}

CLASSWORK
3. Make n the subject of formula if m = 8a + mn^3

Example 4. Given that V^2 – gh = g{sqrt(c^2 + h^2), make h the subject of formula

V^2 – gh = g{sqrt(c^2 + h^2)}
Divide both sides by g
V^2 / g – h = sqrt(c^2 + h^2)
Square both sides
(V^2 / g – h)^2 = c^2 + h^2
Subtract c^2 from both sides

H^2 = (V^2 / g – h)^2 – c^2
Find the square root of both sides
H = sqrt{(V^2 / g – h)^2 – c^2}
H = V^2/g – h – c

CLASSWORK
4. Given r = 2(pi)sqrt(1/t) make t the subject of formula

THE END

m = 8a + mn³

mn³ = m - 8a

............n³ = (m - 8a)/m

n = cubrt(m - 8a)/m

thankyouJesus:
Okay, lets fire on. Could you help with a direct link to physics thread?

no thread yet, because there is no tutor.

Geofavor:

no thread yet, because there is no tutor.

Create one and tag me, I will gladly help where I can.

(B) FACTOR AND REMAINDER THEOREM

To divide polynomials, we write the quotient as a fraction and simplify by using the rules of exponents.

Example 1:

6x²y³ / -2x³y
= -3x^2-3y^3-1
= -3x^-1y²
= - 3y² / x

CLASSWORK
(1) -45r²s⁵t³ / 27r⁶s²t⁸

Example 2:

8x⁵y⁴ + 12x²y⁵ – 16x²y³ / 4x³y⁴
= (8x⁵y⁴ / 4x³y⁴) + (12x²y⁵ / 4x³y⁴) – (16x²y³ / 4x³y⁴)
= (2x^5-3y^4-4) + (3x^2-3y^5-4) – (4x^2-3y^3-4)
= 2x² + 3x^-1y – 4x^-1y^-1
= 2x² + 3y/x – 4/xy

CLASSWORK
(2) 9a³b⁴ + 27a²b⁴ – 18a²b³ / 18a²b⁷

Factor and remainder theorem to be continued shortly...

thankyouJesus:
Create one and tag me, I will gladly help where I can.

alright i will.

(B) FACTOR AND REMAINDER THEOREM cont…

We can also divide polynomials using long division and that is where the remainder theorem comes in.

The remainder theorem states that if a polynomial f(x) is divided by (x – a), the quotient will be a polynomial g(x) of one degree less than the degree of f(x), together with the remainder R still to be divided by (x – a). While this may seem like a herculean task, it is actually very straight forward, easy and interesting.

Example 3: Find the remainder when 2x² + 11x - 30 is divided by x + 7.

We begin by writing the fraction in long division form x + 7 | 2x2 + 11x – 30

2x - 3
------------------------------ (x in 2x² – Put answer on top)
x + 7|2x² + 11x – 30 (Multiply each term of the divisor by 2x)
2x² + 14x (Subtract and bring down the -30)
---------------------------- (x in -3x - Put answer on top)
-3x - 30
-3x - 21
-------------------
-9
The quotient is 2x – 3
Remainder is -9

CLASSWORK
(3) Find the remainder when 2x² – 19x + 35 is divided by x - 7

Note: In cases where the degree power isn't consistent, you could write the missing power of x using co-efficient of 0.

Example 4: Find the remainder when 6x³ - 11 is divided by 2x + 2

3x² - 3x + 3
2x+2 | 6x³ +0x² + 0x - 11
6x³ + 6x²
----------------------------
-6x²
-6x² - 6x
-----------------------------
6x - 11
6x + 6
----------------------------------
-17

Quotient = 3x² - 3x + 3
Remainder = -17

CLASSWORK
(4) Find the quotient and remainder when x⁵ - 32 is divided by x -2

THE END

It is very stressful solving mathematics on the computer. Can i solve on a piece of paper, capture and upload?

Geofavor:

m = 8a + mn³

mn³ = m - 8a

............n³ = (m - 8a)/m

n = cubrt(m - 8a)/m

N = cubrt(1 - 8a/m)

Mathemagician1:
It is very stressful solving mathematics on the computer. Can i solve on a piece of paper, capture and upload?

pls do. But pls make it very large; if not, some of us may not be able to view it.

Mathemagician1:

N = cubrt(1 - 8a/m)

it's the same na.

(C) FACTORIZATION OF POLYNOMIALS

When two of more polynomials are multiplied together, each one is called a factor of the resulting product. For example, the factors of 7(x+2)(x+3) are 7, x+2, x+3.

The process of writing a polynomial as the product of several factors is called factoring.

Example 1: Factorize 3xy² + 6x

A close observation shows that each term contains a common factor of 3x

3xy² + 6x
= 3x(y²) + 3x(2)
We can then use the distributive property to factor out the common factor of 3x.
= 3x(y² + 2)

CLASSWORK
(1) 4a² - 8ab

Example 2: Factorize x²y²z² - xyz
Find the common factor of both term.
= xyz(xyz) - xyz(1)
= xyz(xyz - 1)

CLASSWORK:
(2) a²b²c² + a³b³c³

Factorization of polynomials to be continued shortly....

Download the FREE Pass.ng JAMB CBT APP here --->> https://play.google.com/store/apps/details?id=clickright.educativeApp

(C) FACTORIZATION OF POLYNOMIALS CONT...

A strategy that is often helpful when factoring a polynomial with four or more terms is called grouping.

Example 3: Factorize ax + bx + a + b

Although there is no factor common to all four terms, we can factor x out of the first two terms and 1 out of the last two terms.

ax + bx + a + b
= x(a + b) +1(a + b)
(x + 1)(a + b) Using distributive properties.

CLASSWORK:
(3) x² + xy + 2x + 2y

Factorization of polynomials to be cont...

(C) FACTORIZATION OF POLYNOMIALS CONT...

Another method of factorization is called 'Factoring the difference of two squares'. A binomial that is the difference of two squares of two quantities factors easily. The strategy used is to write the polynomial as the product of two factors. The first factor is the sum of the quantities and the other factor is the difference of the quantities.

Example: Factorize 49x² - 4
Again we observe that each term is a perfect square. Both coefficient and variable.

49x² - 4
= 7²x² - 2²
= (7x)² - 2²
= (7x + 2)(7x - 2)

CLASSWORK
(2) Factorize 9a² - 16b²

Example 5: Factorize 16m⁴ - n⁴

The binomial 16m⁴ - n⁴ can be factored as the difference of two squares.

16m⁴ - n⁴
= (4m²)² - (n²)²
= (4m² + n²)(4m² - n²)

Note that the second factor is the difference of two squares and can still be factored.

(4m² + n²) [(4m² - n²)]
= (4m² + n²) [(2²m² - n²)]
= (4m² + n²)[(2m)² - n²]
= (4m² + n²)(2m + n)(2m - n)

CLASSWORK
(5) Factorize a⁴ - 81b⁴

Example 6: Factorize 18t² - 32
We begin by factoring out common factor of 2, then we can factor the difference of two squares.

18t² - 32
= 2(9t² - 16)
= 2(3²t² - 4²)
= 2(3t + 4)(3t - 4)

CLASSWORK:
(6) Factorize -3x² + 12

(C) FACTORIZATION OF POLYNOMIALS CONT...

Trinomials that are squares of binomials can be factored in many ways but i'll show you perhaps the easiest method to solve this problem.

Example 7: Factorize x² + 3x - 10

We must find two binomials such that x² + 3x - 1. = (x + a)(x + b)
Looking at the equation, we ask, what number can be multiplied to get -10 and when added sums up to +3?

Likely factors of -10 will be:

-10 x 1 = -10
-5 x 2 = -10
10 x -1 = -10
5 x -2 = -10

When added, which will give +3, let's see

-10 + 1 = -9
-5 + 2 = -3
10 + (-1) = +9
5 + (-2) = +3

We've found our factors in 5 and -2. That is a = 5 and b = -2

x² + 3x - 10
x² + 5x -2x - 10
Factorize first two digits and last 2 digits
x(x + 5) -2(x + 5)
Using distributive properties
= (x + 5)(x - 2)

CLASSWORK
(7) Factorize p² - 5p - 6

Example 8: Factorize 2x² - x - 6
Unlike example 7, the coefficient of the first term is 2 and when multiplied by the last term -6 gives -12. We ask. What can be multiplied to get -12 and when added sums up to -1. I find -4 and 3.

2x² - x - 6
= 2x² - 4x + 3x - 6
= 2x(x - 2)+3(x - 2)
= (2x + 3)(x - 2)

CLASSWORK
( 6x² - x - 2

THE END

(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS.

To find the product of two monomials, we multiply the coefficient and then multiply the variables.

Example: Find the product of 3x²y³z and 5xyz².

We proceed as follows.
(3x²y³z)(5xyz²)
= 3*5 x² * x y³ * y z * z²
= 15x³y⁴z³

CLASSWORK
(1) Find the product of 6x²y³ and 3x³y²z

Example 2: To find the product of a monomial and polynomial, we us the distributive property.

See attached for illustration.

CLASSWORK:
(2) 9xy³z(2xy² + 3x²z - 5x²y³)

Example 3: multiply (a) (x+y)(x+y) (b) (x-y)(x-y) (c) (x+y)(x-y)

See attachment for illustration

CLASSWORK:
(3) Multiply (a) (x+2)(x+2) (b) (x-3)(x-3) (c) (x+4)(x-4)

To be cont.

Umartins1:



55_x + 52_x = 77₁₀

5¹5⁰_x + 5¹2⁰_x = 77₁₀

5*x¹ + 5*x⁰ + 5*x¹ + 2*x⁰ = 77

5x+5 + 5x+2= 77

5x+5x+5+2= 77

10x+7=77

10x= 77-7

10x= 70

X= 7.

Pls whr cn I get all dis nairaland code tags in writing stuffs online... Lyk b superscript n subscript stuff u did.

(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS CONT...

Example 4 and 5: Please see attached.

CLASSWORK:
(5) (2r^-3 - 7n)(3r² - 2n^-2)

To be cont...

(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS CONT...

Example 6: see attached.

(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS CONT...

Example 7: See attached.

We've already covered division of polynomials under factor and remainder theorem.

THE END

Thank God I'm in the right thread. May God bless you for us, @Mr. Mathemagician

Mathemagician1:
(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS CONT...

Example 7: See attached.

We've already covered division of polynomials under factor and remainder theorem.

THE END

Can I go on with my topic?

mathefaro:
Thank God I'm in the right thread. May God bless you for us, @Mr. Mathemagician

See as I dey look you.

thankyouJesus:
Can I go on with my topic?See as I dey look you.

Sir please proceed. You don't have to wait, just put your topic on the header and fire on.

thankyouJesus:
See as I dey look you.

Abeg, my oga, na low-key I dey I just dey o. Make I learn small from the bosses like you

Mathemagician1:


Sir please proceed. You don't have to wait, just put your topic on the header and fire on.

Okay o

mathefaro:
Abeg, my oga, na low-key I dey I just dey o. Make I learn small from the bosses like you

Good evening boss, dey whyne me o

[quote author=thankyouJesus post=40882311]Good evening boss, dey whyne me o

I hail you bro. You're doing a very good job here

Pardon me to start again.

thankyouJesus:
[b]Good evening all, as per my obligation, today will be "our" first class. "We" will be considering differentiation.
Note 1:
1. dy/dx can also be represented as y'.
2. d²y/dx² can also be represented as y", till infinity.
3. d/dx means "we" are differentiating with respect to x, d/du means we are differentiating with respect to . . . . . . . . . . . .very good, u.
4. wrt means with respect to.
DIFFERENTIATION.
Differentiation means rate of change.
General formula:
if y = axⁿ,
a is the co-efficient of x.
n is the power of x.
Therefore, dy/dx = power "times" co-effecient of x "times" the power of x "minus" 1.
Mathematically,
y = axⁿ,
dy/dx = nax^{n - 1}.
-----------------------------
Example 1. Find y' [it also means dy/dx] if y = 2x.
Solution:
the co-efficient of x is 2.
The power of x is 1.
Therefore,
y = 2x
dy/dx = (1)(2)x^1-1
dy/dx = 2x⁰
dy/dx = 2(1)
because x⁰ is 1.
dy/dx = 2.
-----------------------------.
Example 2: find y' if y = 3x⁴.
Solution:
y = 3x⁴
y = 12x³
-----------------------------.
Example 3: find y' if y = x
.
.
.
.
.
.
.
y' = 1.
Very good, because,
y = x
the co-efficient of x is 1.
The power of x is 1.
Therefore,
y = x
y' = (1)(1)x^1-1
y' = 1x⁰
y' = 1(1)
y' = 1.
-----------------------------.
Example 4: find y' if y = 2.
Solution:
y = 2,
the question can also be changed into
y = 2(1)
it can be further changed into
y = 2(x⁰)
since,
y = 2x⁰
y' = (0)(2)x^0-1
y' = 0x^-1
y' = 0.
Note 2: The derivative of a constant is 0.
i.e
d/dx (7) = 0
d/dx (10) = 0, but,
d/dx (7x) = 7
d/dx (10x) = 10.
-----------------------------.
Summary:
1. Differentiation means rate of change.
2. If y = axⁿ, then dy/dx = nax^n-1.
3. The differentiation of constant values is 0. E.g,
y = 6,
dy/dx = 0,
y = 1000000000
dy/dx = 0.
-----------------------------.
Assignment: find dy/dx if y =
1. 20x
2. -50x³
3. -25x^-2
4. 30/x², [Hint: 1/x² is x^-2.

Any question?

Good night class.
23rd March, 2015.
21:35:45[/b]

www.nairaland.com/thankyoujesus/posts/16

(E) ROOTS OF POLYNOMIALS 
‎
Before we begin this topic please make sure you have previous knowledge of complex numbers and complex conjugates. Also read up how to divide polynomials using synthetic division.‎
‎
There are three important theorems relating to the roots of a polynomial. They are:‎

(1) Every‎ ‎n‎th degree polynomial has exactly n linear factors‎.

(2) A polynomial equation of degree (‎n) ‎has exactly n roots.

(3) If (x − r) is a factor of a polynomial, then x = r is a root of the associated polynomial equation.

Example 1: Fine the possible rational roots of the polynomial f(x) = 2x² + 6x.

‎The polynomial f(x) = 2x² + 6x is a second degree polynomial‎ (since the highest power of x ‎that appears is 2). 
‎
According to the theorems, since the highest degree is 2, therefore the polynomial has 2 factors and 2 roots. 

We can find the root of this polynomial by factoring.‎

Please see attached for solution.

Example 2: Find the possible rational roots of 2x² + 4x - 6.

Please see attached for solution. 

CLASSWORK:
(2) Find the possible rational roots of g(x) = 10x² - 8x -2

To be cont...‎