# Grade 10 Quadratic Unit

### The Basics of the Algebraic portion

## Introduction

## Factoring

ab+ac=a(b+c)

In this equation a is the common factor

Polynomials are not considered to be fully factored until the greatest common factor(G.C.F) has been factored out

Let us consider the following expression, 8x-12y

Now both 8x and 12y can be divided by 4 as 4 is thre G.C.F.

we would therefore start our factored expression off with 4 and then continue the dividing process

8x divided by 4 would be 2x and 12y divided by 4 would be 3y. We now have the information needed to make our factored form.

Therefore the new equation would be

4(2x-3y).

Please view the example as explained by Khan Academy below. The first 4 minutes will suffice in explaining the very basics of factoring.

## Expanding

Let us consider the following:

(x+a)(x+b)

if we expand this by multiplying each term in the first bracket by the 2nd we get:

(x)(x)+ bx+ax+ab

After we expand our next step is to to simplify.

x^2 +(a+b)x+ab

If we were to apply the distributive property to an exponent it would work like this in theory.

(x-5)^2

we would first have to make 2 groups as (x-5) is all to the exponent of 2.

Then we would get (x-5)(x-5) and then work on expanding from there.

With that accomplished we should get x^2 -5x-5x+25

After simplifying(adding the x's)

we will get x^2-10x+25

Therefore x^2-10x+25 would be our final answer.

## Solving

*ax^*2 +

*bx*+

*c*= 0. For the purpose of explanation we will solve 1x^2 +3x-4=0 using the quadratic formula.

Our first step will be to write out the formula of x= -b+/- √b^2-4ac divided over 2a.

Our next job is to start substituting the values we received from the equation. In this example, 1 would be a, 3 would be b and -4 would be c. when we substitute these values into the formula we get the following: -3 +/- √ 3^2-4(1)(-4) all divided by 2(1)

Most of the work will come from the former step as the only thing left to do is to use the BEDMAS order(brackets,exponents,multiply,addition,subtraction) to break it down into our 2 x values.

With that said it will break down to -3 +/- √ 9+16 all divided by 2. The 9 came from solving the exponents the 2nd part of the BEDMAS order

when we add 9 and 16 the equation will become x=-3+√25 all divided by 2

From here we find the square root of 25 before we proceed further. The square root of 25 is 5 therefore √25=5

Now because of the + and - sign just after the b value we have 2 answers

one will be positive and the other negative

x= (-3+5)/2 and (-3-5)/2

we must solve the brackets and divide by 2 to receive our two answers

They will be x=-4 and x=1

Therefore the x intercepts will be -4 and 1

## Completing the Square

The equation is currently in standard form, if we wished to find the vertex form(so we could figure out how to graph it later on) we would have to complete the square

The first step would be to** block off the very first 2 terms** so we would get

y=(x^2+6x)+11

Our Next step would be to factor out the A but since the A value is equal to 1 we may skip this step

Our 3rd step would be to divide the middle term(6x in this example) by 2 and then square it

It would look like (6/2)^2=9

After getting the answer we plug it back into the equation to get y=(x^2+6x+9-9)+11

Lastly we take the negative out of the bracket by multiplying it by the A value outside of the bracket(1)

(x^2+6x+9)+11-9

After taking out the negative we square the equation to get

(x+3)^2+2