Let f (x)=6x^3-7x^2-9x-2 and substitute these values to x and you will find f (2)=0. This means that x=2 is one of the roots and thus f (x) is divisivle by (x-2).

How do you find the radius of a circle with the equation #x^2 + y^2 - 6x - 12y + 36 = 0#? How do you find the radius of a circle with the equation #x^2 - 8x + y^2 - 4y – 5 = 0#?

f (x) has zeros +-sqrt (5) and -2 Given: f (x) = 3x^3+6x^2-15x-30 Note that all of the terms are divisible by 3. In addition, the ratio of the first and second terms is the same as that of the third …

Let "mass of X" = x and "mass of O" = y. y/ (x + y) = 0.6 y = 0.6x + 0.6y 0.4y = 0.6x y = (0.6x)/0.4 = 1.5x Let "mass of x = 32 u". Then "mass of O = 1.5 × 32 u = 48 u". ulbb ("Element"color …

Jan 29, 2017 · 6x^2+x-1 = (2x+1) (3x-1) Here are a couple of methods (in no particular order): Method 1 Note that: (ax+1) (bx-1) = abx^2+ (b-a)x-1 Comparing with: 6x^2+x-1 we want to find …

0, 1, 3 To solve this, we set the equation equal to 0 and factor it. First, we can factor out 6x to get 6x(x^2-4x+3)=0. Next we can factor out the area in parentheses. To do this, we need to find …

How do you use the important points to sketch the graph of y = x2 − 6x + 1? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs

We're dividing a single term, #6x^3#, into a polynomial, so we can take advantage of the following fact which allows us to effectively divide one term into other terms:

x=9 So we have: x-1=sqrt (6x+10) Let's try to remove the square root sign. We square both sides. (x-1)^2= (sqrt (6x+10))^2 x^2-2x+1=6x+10 We see that we can form a quadratic equation. …

May 19, 2018 · 59 degrees Since ABCD is a cyclic quadrilateral, then angle ABC + angle ADC = 180 (opposite angles in cyclic quadrilateral are equal to 180 degrees) 2x+3+4x+3=180 6x ...