3x²-15x+12=0 1.-Find the GCF 2.-then place two sets of parenthesis 3.-factor the front term 4.-then find two factors of the last term that add up to the second term. 5.-find the solution by equaling every part to zero 6.Then solve for 0 1.GCF is 3. 3( x²-5x+4)=0 2.( ) ( ) 3.(x ) (x ) 4.(x-4) (x-1) 5.3=0, x-4=0, x-1=0 6.(Because 3=0 is not true it is cancelled out.) x-4=0 x-1= x=4 x=1 Solutions= 4,1

 x²+3x-28=0  ( ) ( )  (x ) (x )  7*(-4)= (-4)= 3  (x+7) (x-4)  x+7=0 x-4=0  Solutions= -7,4

 ( ) ( )  (6x ) (6x )  -10*(-9)= (-9)= -19  (6x-10) (6x-9)  (3x-5) (2x-3)  Solutions=5/3, 3/2

 Is a function that has the form of ax²+bx+c. It’s graph is a parabola. The difference with a linear function is that a linear function has a form of y=mx+b and a quadratic function has x².

LINEAR FUNCTIONS QUADRATIC FUNCTIONS y=5x-3 y= 3x+1 y= 4x y= 10x-3 y=3x²+4 y= x²-2x+7 y= 2x²-3x+9 y= x²+7x-10

STEPSGraph y=x²-4 1.Find the y intecept that is c. 2.Find the vertex using –b/2a 3.Decide if the graph opens up or down. a>0 opens up, a<0 opens down 4.Find appropriate values for y and do a table. NOTE: if the graph opens up, the vertex is a minimum point and if the graph opens down the vertex is a maximum point. 1.y intercept= -4 2.Vertex= -b/2a = -0/2 = 0 3.Opens up 4.

 y-intecept =0  vertex= -b/2a= 0/-2= 0  Opens down 

 y-intercept= 6  Vertex= -b/2a= 4/4= 1  Opens up

 You have to graph the function and look for the x-intercepts. The x-intercepts or the functions are the solution.  Also when you have y=ax²+bx+c; › a changes steepness of the parabola › b moves right to left in the x axis, positive moves to the left and negative to the right › C moves vertex up or down in the y axis. Positive moves up, and negative moves down.



STEPS5x²=20 1.First leave x² alone 2.When an expression is a perfect square you can find the squared root of both sides and cancel the squared root of one side. 1. x²=4 2.x= + or - 2

  x= + or - 4 X²+8=28 Isolate X² and subtract 8 to both sides X²=

StepsX²+3x=40 1. leave zero at one side of the equation. 2. factor the polynomial ( follow the same steps as the first slide) 3.The at the end find the solution 1.X²+3x-40=0 2.(x+8) (x-5) =0 3.Solution= -8,5

 (x-2) (x+2) = 0  X-2=0 x+2= x=2 x=-2 Solutions= 2,-2

 Subtract 11x from both sides  2x²-11x+12=0  Multiply a and c  2x²-11x+24=0  (2x-8) (2x-3)  Solutions= 3/2, 4

STEPS3X²-2x=9 1.Get x²=1 2. isolate c 3.Find b, divide b/2 and then square it (b/2)² 4.Find the squared root of both sides. 5.Isolate x buy adding or subtracting in both sides. 1.x²-2/3x= (2/3)/2²= 1/9 (Add 1/9 to both sides.)

 Subtract 16 from both sides  x²-10x=-16  (10x/2)²= 25  x²-10x(+25)=-16(+25)… parenthesis in this case are not multiplying.

 x²+2x=-63  (2/2)²=1

 Leave zero at one side. Identify the coefficients a, b, c. Then use the formula. The discriminant is the expression inside the squared root.  D>0 positive and has two solutions  D=0 one solution  D<0 negative and is no solution because the squared root of a negative number does not exist.

 A=1 B=-8 C=12  Plug in what you know to the equation

