By: Suhasee

Learning Goal 1

By the end of this unit I should be able to:

Expand and Simplify Binomials

Expanding the brackets involves removing the brackets from an expression by multiplying out the brackets. This is achieved by multiplying every term inside the bracket by the term outside the bracket. When multiplying out the double brackets, every term in the first pair of brackets must be multiplied by each term in the second.

Ex. Single pair of brackets:



Learning Goal 2

By the end of this unit I should be able to:

Graph the parabola

In order for us to graph the parabola in vertex we would need to know the x-intercepts, the y-intercepts and the vertex. Once we have these three points we are then able to graph the parabola using the coordinates for the graph.

At first when we do this we should set y as 0 and solve and factor in order to find the 2 or 1 x-intercepts. Once, we do that we would be able to find the axis of symmetry but adding the 2 x values together and dividing it by 2. Then whatever number that we get we should sub in that with the normal equation . Once we do that we are able to find the y value. That would give you the vertex.

Summary Of The Unit

Expanding and Simplifying Polyinomials

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Multiplying Binomials -- Use the distributive property to multiply binomials.​

Factoring Polynomials

Using The GCF

No matter how many terms a polynomial has, you always want to check for a greatest common factor (GCF) first. If the polynomial has a GCF, factoring the rest of the polynomial is much easier because once you factor out the GCF, the remaining terms will be less cumbersome. If the GCF includes a variable, your job becomes even easier.

When solving for x in a polynomial equation, if you forget to factor out the GCF, you may miss a solution, and that could mix you up in more ways than one! Without that solution, you could end up with an incorrect graph for your polynomial. And then all your work would be for nothing!

To factor the polynomial 6x4 – 12x3 + 4x2, for example, follow these steps:

  1. Break down every term into prime factors. This step expands the original expression to

  2. Look for factors that appear in every single term to determine the GCF.

    In this example, you can see one 2 and two x’s in every term:

    The GCF here is 2x2.

  3. Factor the GCF out from every term in front of parentheses and group the remnants inside the parentheses.

    You now have

  4. Multiply each term to simplify.

    The simplified form of the expression you find in Step 3 is 2x2(3x2 – 6x + 2).

    To see if you factored correctly, distribute the GCF and see if you obtain your original polynomial. If you multiply the 2x2 inside the parentheses, you get 6x4 – 12x3 + 4x2. You can now say with confidence that 2x2 is the GCF.

Factoring By Grouping

What is Factoring by Grouping 1 - Algebra Help - IGCSE GCSE Maths

Simple and Complex Trinomials

Simple Trinomials: Simple Trinomials are when the value of "a" is equal to 1. All you have to do is find the product and the sum. Since a = 1 then we have to find two numbers whose product is c and two numbers whose sum is b.

For example: x^2+12x+27

For this we would need to find two number whose product equal to 27 and the sum of those two numbers equal to 12. For this specific example the two numbers would be 9 and 3. Therefore, in order to factor this we would write x^2+12x+27 = (x+3)(x+9)

Complex Trinomials: Complex Trinomials are when the value of "a" is greater than or less than 1. We have to find two numbers whose product is and c and those two numbers whose sum is equal to b.

For example: 3x^2+7xy+2y^2

For this we would need to identify the "a", "b", and "c" values, which are a=3, b=7, c=2

Now you would multiple the "a" and "c" values which would be 3(2) = 6. So, two numbers have to have the product as 6 and the sum of them should be 7. The factors for this equation is 6 & 1. This is so because 6(1) = 6, 6+1=7. Now you would replace 7xy with 6xy and xy.


Now we would just common factor it so:

3x(x+2y) +y(x+2y)


Difference of Squares

You can factor the difference of squares as a 2 - b 2 = (a+b)(a-b) Now we do the exact opposite of factoring, so the pattern a 2 - b 2 = (a+b)(a-b). This is known as the difference of squares, both a and b must be perfect squares ( if you take the square root, it is a whole number.)

For Example: x^2 - 100

Now you would square both the first and second term. This would result in x and 10. Therefore, the answer to this expression would be (x + 10)(x - 10).

Perfect Square Trinomials

In a perfect square trinomial the first and middle terms are perfect squares and the middle term is twice the product of the square roots of the first and last term.

For Example: x^2 + 12x + 36

Both x^2 and 36 are perfect squares and 12x is twice the product of x and 6.

Since all the signs are positive the pattern is a=x and b=6. Therefore the answer to this is (x+6)^2 or (x+6)(x+6).

Factoring to Determine X

Using the 3 point method we are able to determine the x-intercepts in the equation. The first step to do that is is to replace y with 0 and then begin to common factor. For this there are 4 different cases.

The first case it would be to replace y with 0 and then to common factor, by doing this the x-intercepts would be found and one of them would always be 0. In order to find out the vertex from that point you would take both the x-intercepts and add them together, and then you would divide them by 2. The amount that you get, you should replace it with the variable x and by doing that it would help determine the y value. By doing that it will help you find the vertex of the quadratic.

The second case is quite similar you would be set y=0 and then common factor. But since this case it only with perfect square trinomials you would factor the perfect square and determine the x-intercept. And then you would sub the x-intercept into the original equation.

The third case is about difference of squares and if in that case then you woul common factor first and then factor the difference of squares and then state the two x-intercepts and then lastly determine the vertex.

The fourth and final case is when you are not able factor fully and you can only partial factor. You would do this by common factoring quadratic and linear term only. After you do this can are able to determine the y-intercepts and the reflected points from the partial factor. After finding out the y-intercept and reflected point to determine the vertex.

Ex: y= x^2 + 5x - 8

y= x(x+5) - 8

x=0 x=-5

Therefore, y=-8 and -8; so the x-intercept is (0, -8) and reflected point is (-5,-8). And in order to determine AOS you would 0+(-5)/2 which equals to -2.5. Now you would sub in -2.5 in x in the original equation.

y= (-2.5)^2 +5(-2.5)-8

y= -14.25

Word Problem

The face of a Canadian $20 dollar bill has a area that is represented by 10x^2+9x-40.

a) Find the binomials that represent the dimensions.

The bionomials are 25x and -16x.




Therefore the dimensions are l=(5x-8) and w=(2x+5)

b) What if x=32? What would the dimensions be?

(5x-8) (2x+5)

(5(32)-8) (2(32)+5)

160-8 64+5

152 69

Therefore, the length is 152mm and the width is 69mm.