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Factorising quadratics with rational roots

October 31, 2010 Leave a comment

We are going to consider quadratics of the form ax^2+bx+c where we can put them in the form (mx+n)(px+q)

A prerequisite of this post is to have read Factorising into a single bracket
Also it should be noted that it is not possible to factorise all quadratics this way, since they do not all have rational roots and some can not be factorised at all.

Step by step instructions

  1. Multiply a by c (the x^2 coefficient and the constant)
  2. Find all the factor pairs to make this number
  3. Total the pairs and see which pair equals the x coefficient
  4. Split the x term according to the chosen pair
  5. Factorise into single brackets the first and second pair of terms.  Make sure that you use the sign of the third term
  6. Finally factorise the two resulting terms, where the common factor is the bracket
  7. If the brackets are different then the expression does not have rational roots and another method, like completing the square must be used

Examples

Factorise 6x^2+11x+4

6 \times 4 = 24

Factor pairs of 24

1 & 24, -1 & -24, 2& 12, -2 & -12, 3 & 8, -3 & -8, 4 & 6, -4 & -6

The pair that have a total of 11 are 3 & 8

\therefore we change the 11x \rightarrow 3x + 8x

So the expression becomes 6x^2+3x+8x+4

Factorising each pair gives 3x(2x+1)+4(2x+1)

Now factorise each term (common factor (2x+1)

We get (2x+1)(3x+4)

(Note: the brackets could be in either order)


Factorise 3x^2+10x-8

3 \times -8 = -24

Factor pairs of -24

1 & -24, -1 & 24, 2& -12, -2 & 12, 3 & -8, -3 & 8, 4 & -6, -4 & 6

The pair that have a total of 10 are -2 & 12

\therefore we change the 10x \rightarrow 12x - 2x

So the expression becomes 3x^2+12x-2x-8

Factorising each pair gives 3x(x+4)-2(x+4).  (Note we factor out -2, we take the sign of the third term)

Now factorise each term (common factor (x+4)

We get (x+4)(3x-2)

Categories: Algebra

Factorising quadratics with integer roots

October 31, 2010 Leave a comment

The general form of a quadratic is ax^2+bx+c

We are going to consider cases where a=1

Consider (x+\alpha)(x+\beta)

If we expand the brackets we get x^2+\alpha x + \beta x + \alpha \beta

This simplifies to x^2+(\alpha + \beta)x + \alpha \beta

It can be seen from above that the sum of \alpha and \beta is equal to the coefficient of x

and that the product of \alpha and \beta is equal to the value of the constant.

So to factorise a quadratic first:

  • Find all the pairs of integers, positive and negative, that multiply together to make the constant value
  • See if one of the pairs if added instead can equal the coefficient of x
  • If such a pair can be found use them to replace \alpha and \beta in the formula (x+\alpha)(x+\beta)
  • If a pair can not be found then a different method is needed

Eamples

Factorise x^2+8x+15

First lets find the pairs of numbers that when multiplied together make +15.

1 15
-1 -15
3 5
-3 -5

Now see if any pair when added together we get the value +8

1 15 16
-1 -15 -16
3 5 8
-3 -5 -8

Since 3 & 5 multiply together to make 15 and add together to make 8 these are out \alpha and \beta

So x^2+8x+15=(x+3)(x+5)

(Note that I tried to be systematic when finding the factor pairs which equalled 15.  Also note that it does not matter if you give the factorised expression as  (x+3)(x+5) or (x+5)(x+3)


Factorise x^2-7x+12

 

The factor pairs

1 12
-1 -12
2 6
-2 -6
3 4
-3 -4

Now total each pair

1 12 13
-1 -12 -13
2 6 8
-2 -6 -8
3 4 7
-3 -4 -7

Since -3 & -4 multiply together to make 12 and add together to make -7 these are out \alpha and \beta

So x^2-7x+12=(x-3)(x-4)


Factorise x^2+3x-10

 

The factor pairs

1 -10
-1 10
2 -5
-2 5

Now total each pair

1 -10 -9
-1 10 9
2 -5 -3
-2 5 3

Since -2 & 5 multiply together to make -10 and add together to make 3 these are out \alpha and \beta

So x^2+3x-10=(x-2)(x+5)


Factorise x^2-4x-21

 

The factor pairs

1 -21
-1 21
3 -7
-3 7

Now total each pair

1 -21 -20
-1 21 21
3 -7 -4
-3 7 4

Since 3 & -7 multiply together to make -21 and add together to make -4 these are out \alpha and \beta

So x^2-7x-21=(x+3)(x-7)


Factorise x^2-2x-7

 

The factor pairs

1 -7
-1 7

Now total each pair

1 -7 -6
-1 7 6

Since no pair has a total of -2 that we required and because we have been systematic we know we have considered all possible pairs, we can assume that either this expression does not factorise or that a different method is required.


Categories: Algebra

Factorising into a single bracket

October 30, 2010 Leave a comment

When factorising into a single bracket we divide out the highest common factor (H.C.F.) of each term.  This H.C.F. becomes the brackets multiplier and the result of the division becomes the contents of the bracket.

Examples

3x+6 \rightarrow 3(x+2)

Here we can see that 3 divides into both terms, but x does not so the most we can divide out is 3.

3x^2-5x \rightarrow x(3x-5)

With this expression there was no number that could be taken out of all terms, but there was an x in both

6x^3-9xy \rightarrow 3x(2x^2-3y)

This time we can divide both terms by 3 and x.  We don’t have to check for y since it is not in the first term

4x^2y+8xy^2-6xy \rightarrow 2xy(2x+4y-3)

Here we can divide out 2, x and y from all three terms.  The number of terms does not effect the problem provided that the factor that is divided out is common to all three.

4(p+q)+(p+q)^2 \rightarrow (p+q)(4+(p+q))=(p+q)(4+p+q)

In the strange example above, both terms had a factor (p+q), therefore we could divide this out.  Make sure you can follow this example, we will be using this idea again in  a later post.

3(x+y)^2+6x(x+y)^3 \rightarrow 3(x+y)^2(1+2x(x+y))=3(x+y)^2(1+2x^2+2xy))

Categories: Algebra

Multiplying out brackets

October 29, 2010 Leave a comment

Multiplying a bracket by a single term

  • We must take account of the sign of the term
  • Each and every term in the bracket is multiplied by the term out side

Examples

3(x+5)=3x+15

In the example above we multiplied the x and 5 by 3

This is illustrated in the diagram where one side is of length 3 and the other is x+5.  The area is the required answer.

x(x-4)=x^2-4x

In the example above we multiplied the x and -4 by x

-2(y-3)=-2y+6

In the example above we multiplied the y and -3 by -2

4+2(x-5)=4+2x-10=-6+2x

In the example above the +2 is the only entity that is multiplying the contents of the bracket

Multiplying two brackets together

Consider the problem (x+3)(x+6)

If we look at the illustration above we can see that each element of the first bracket is multiplied by each element of the second bracket.

The result is that (x+3)(x+6)=x^2+6x+3x+18=x^2+9x+18

To help make sure that we multiply every thing in the first bracket by everything in the second bracket (provided that there are exactly two in both) we can use the mnemonic FOIL

  1. F multiply the first element in each bracket together
  2. O multiply the outside elements of the expression together
  3. I multiply the inside elements of the expression together
  4. L multiply the last element in each bracket together

Examples

(x+2)(x-5)=x^2-5x+2x-10=x^2-3x-10

Take notice of the order that I reached my terms (I used FOIL) and the fact that I simplified the expression afterwards.

(2t-2)(3t-5)=6t^2-10t-6t+10=6t^2-16t+10

Notice that multiply by the directed (signed) value not just the positive value

Multiplying multiple different sized brackets together

  • When you multiply brackets together make sure that all elements the first bracket are multiplied by all elements of the second
  • Multiply a pair of brackets together first, if there are more than two, to reduce the number of brackets by one
  • Keep on doing this until all brackets have be multiplied out

Example

(2x+3)(x-1)(x+2)

multiply the first two brackets together

=(2x^2-2x+3x-3)(x+2)

Simplify the new first bracket

=(2x^2+x-3)(x+2)

Multiply out the brackets.  Because there are three terms in the first bracket and two in the second we would expect to get 2 \times 3=6 terms in the expansion

=2x^3+4x^2+x^2+2x-3x-6

Simplify the answer

=2x^3+5x^2-x-6

Questions

Categories: Algebra

Lines intersecting circles

October 29, 2010 Leave a comment

When a line intersects a circle it will either have one point of contact in which case it is a tangent to the circle or it will cut it in two places.

To find the points of contact between a line and a circle

  1. If needed, rearange the equation of the line to make x or y the subject
  2. Substitute into the circle equation and solve the resulting quadratic
  3. Substitute the solutions into the equation of the line to get the required coordinates

Example

Find where the line y=2x+1 intersects with the circle (x+2)^2+(y-1)^2=5

When we substitute y into the circle we get (x+2)^2 + ((2x+1)-1)^2=5

\therefore (x+2)^2+(2x)^2=5 \Rightarrow x^2+4x+4+4x^2=5 \Rightarrow 5x^2+4x-1=0

\therefore (5x-1)(x+1)=0 \Rightarrow x=\frac{1}{5} or x = -1

If x=\frac{1}{5}, y=\frac{7}{5}

If x=-1, y=-1

\therefore the points of intersection are (\frac{1}{5}, \frac{7}{5}) and (-1,-1)

Categories: Graphing

Where two lines intersect

October 29, 2010 Leave a comment

Using the substitution method to find where two lines intersect

  1. Rearrange one of the equations to make either x or y the subject
  2. Substitute for either x or y depending on which one you made the subject
  3. Solve the resulting equation
  4. Substitute this value into one of the original equations to find the value of the other variable

Example

Find the coordinates where the lines y=3x+5 and 3x+2y=28 intersect

Since the first equation already has y as the subject we will substitute this into the second

\therefore 3x+2(3x+5)=28 \Rightarrow 9x+10=28 \Rightarrow x=2

We now substitute this back into the first (on this occassion it is the easiest option)

and get y=3 \times 2+5=11

\therefore the point of intersection in (2,11)

Categories: Graphing

Finding the turning points of a polynomial

October 29, 2010 Leave a comment

The turning points are sometimes called the stationary points or maximums and minimums.

Lets look at what happens to the gradient at a turning point

  • If we consider the maximum, you can see that before it has a positive gradient and after it has a negative gradient.  During this transition it has a gradient of zero.  So at a maximum \frac{dy}{dx}=0
  • If we consider the minimum, you can see that before it has a negative gradient and after it has a positive gradient.  During this transition it has a gradient of zero.  So at a minimum \frac{dy}{dx}=0
  • So we can see that at a turning point (maximum or minimum) \frac{dy}{dx}=0

Examples

Find the maximum of the curve y=15-2x-x^2

  1. We need to find the gradient function \frac{dy}{dx}
  2. We let \frac{dy}{dx}=0, because this is when the maximum will occur
  3. We solve the equation to find the value of x where the maximum occurs
  4. We substitute this value in the original equation to get the value for y

\frac{dy}{dx}=-2-2x

Let \frac{dy}{dx}=0 \Rightarrow -2-2x=0 \Rightarrow x=-1

If x=-1 \Rightarrow y = 15-2(-1)-(-1)^2 = 16

\therefore the maximum value is y=16

Find the coordinates of the turning points of y=x^3-6x^2+9x

  1. We need to find the gradient function \frac{dy}{dx}
  2. We let \frac{dy}{dx}=0, because this is when the turning points will occur
  3. We solve the equation to find the values of x where the turning points occurs
  4. We substitute the values into the original equation to get the values for y

\frac{dy}{dx}=3x^2-12x+9

Let \frac{dy}{dx}=0 \Rightarrow 3x^2-12x+9=0 \Rightarrow 3(x^2-4x+3)=0 \Rightarrow 3(x-1)(x-3)=0

So x=1 of x=3

If x=1 then y=4 and if x=3 then y=-18

So the coordinates of the turning points are (1,4) and (3,-18)

Given that the curve y=x^2+ax-5 has a stationary point at x = 4 find the value of a

  1. We need to find the gradient function \frac{dy}{dx}
  2. We let \frac{dy}{dx}=0, because this is when the turning points will occur and we will let x = 4
  3. We will solve the resulting equation to find the value of a

\frac{dy}{dx}=2x+a

when x=4, \frac{dy}{dx}=0 \Rightarrow 0=8+a \Rightarrow a=-8

 Find the equation of the tangent to the curve y=2x^3+3x-5 at x=1

  1. We need to find the gradient function \frac{dy}{dx}, because to get the equation of a line we need the gradient and a point it passes through
  2. Substitute x=1 into the equation for y to get the required point the line passes through
  3. Substitute x=1 into the equation for \frac{dy}{dx} to get the gradient of the line
  4. Use y-y_1=m(x-x_1)  to get the equation of the line

\frac{dy}{dx}=6x^2+3

At x=1, y=0

At x=1, \frac{dy}{dx}=9

So substituting into y-y_1=m(x-x_1) gives y=9(x-1)

Categories: Differentiation