Learning goals

• I will know what each variable represents in the equation y = a(x-r)(x-s)
• I will learn about the different types of factoring
• I will learn how to find the x- intercepts by factoring

Summary of unit

• Factored form uses the equation y=a(x-r)(x-s)
• The value of a gives you the direction and opening of a parabola
• The values of r and s give you the x-intercepts
• You can find the axis of symmetry,(A0S), by: x=(r+s)÷2. substitute this x value into the original equation to find the y value of the vertex
• To find the y-intercept, set x=0 and solve for y ,Solve using the factors

There are different types of factoring, but it is not always possible to factor an equation

Greatest common factoring: Greatest common factoring involves taking the greatest common factor (number and/ or variable) out of the equation to simplify it.

Ex. 28c²-21c

= 7c (4c-3) In this equation 7c is the greatest common factor of 28c² and 21c

simple trinomials: In simple trinomials, the equation in standard form is x²+bx+c. This equation in factored form looks like (x+r)(x+s). r+s= b and (r)(s)= c. The value for x² in simple trinomials is always 1 and can never be 0. To factor, you have to find the product of the last term which is equal to the sum of the middle term.

Ex. x²+12x+27

The factors for the last term (27) are 1 and 27, 3 and 9. out of these factors,3 and 9 add up to the middle term (12). so the factors for the equation x²+12x+27 would be (x+3)(x+9).

• If b is negative and c is positive (x²-29x+28), both r and s are negative. eg.(x-1)(x-28)
• If c is negative, ONE of r or s is negative (x²+3x-18). the factors are (x-3)(x+6)
• If b and c are both negative, ONE of r or s is negative (x²-5x-24) the factors would be (x-8)(x+3)

Complex trinomial factoring: In complex trinomials, the value of a can never be 0 or 1 from the equation ax²+bx+c . To factor this equation, we have to find two integers whose product is a×c and whose sum is b. we also have to break up the middle term (b) by factoring.

Ex. 3x²+14x+8 The product of 3 and 8 is 24. 24÷2= 12. 12+2=14 (middle term)

=3x²+12x+2x+8

= 3x(x+4)+2(x+4) we count x+4 as one term

= (3x+2) (x+4)

Ex. 16x²+26x-12 In this case we need to take the common factor out first

= 2(8x²+13x-6)

=2 (8x²+16x-3x-6)

= 2[8x(x+2)-3(x+2)]

= 2 (8x-3) (x+2)

Difference of squares: Is only possible in binomials. In difference of squares the formula used is (a+b) (a-b) . Both a and b must be perfect squares ( if you take the square root is it a whole number). The sign that separates the first term and the second term must be a negative sign.

Ex. x²-100

= √x √100

= (x+10) (x-10)

Perfect square trinomials: The trinomial that results from squaring a binomial is called a perfect square trinomial. you can factor a perfect square trionomial as a²+2ab+b² or

a²-2ab-b². The middle term should be twice the product of the square roots of the first and second term. 2(√a × √b)= 2abnd term. 2(√a × √b)= 2ab

Ex. x²+6x+9

= 2(√ x² × √9)

= 2(3x)

=6x Since twice the square root of the first term and the second term is equal to 6x, the factor this perfect square trinomial is (x+3)²

Determining the x- intercepts: set your equation to y=0. Then find individual factors and set them to 0 and solve for x.

Ex. y=-x²-5x

0= (-x) (x+5)

-x=0 x+5=0

x=0 x=-5

Word problem

The path of a toy rocket is defined by the relation y=-3x²+x+4, where x is the horizontal distance in meters travelled, and y is the height in meters above the ground.

a) What was the initial height of the rocket?

b) Determine the zeros of the relation.

c) How far has the rocket travelled horizontally when it lands on the ground?

d) What is the maximum height of the rocket above the ground, to the nearest hundredth of a meter?

a) y=-3x²+x+4

y=-3(0)²+(0)+4

y=4 The initial height of the rocket is 4 meters (0,4)

b) y=-3x²+x+4

0=-1(3x²-x-4)

-1(3x²-4x+3x-4)

-1[x(3x-4)+1(3x-4)]

-1(x+1) (3x-4)

x+1= 0 3x-4=0, 3x ÷3=4÷3

x= -1 x= 1.3

The zeros of this relation are x=-1 and x=1.3

c) The rocket has travelled 2.3 meters [ from point (-1,0) to point (1.3,0) on the graph]

d) Aos =( -1+1.3) ÷2 = 0.15

y=-3x²+x+4

y=-3(0.15)²+(0.15)+4

y=-0.0675+0.15+4

y=0.0825+4

y= 4.0825

The maximum height of the rocket is 4.08 meters at 0.15 meters (0.15,4.08).