d) determine the dominant term e) determine the degree f) determine the end behaviour

Q: (x² +5x-4) + (2x²+2x+3)

A: x2+5x-4+2x2+2x+3

Q: Q2 (2x - 1)(x+2)

A: Let,the given expression is

Q: If Q(x) = 7x - 1, find Q(- 9). Q(- 9) =

A:

Q: 5x-2(2x5+x2)

A: Consider given expression as K=5x-22x5+x2

Q: 2(x -3) – (4x + 5)

A:

Q: 4(2x+3)-3(5-x)=2(x+30)

A:

Q: 2(4x-3)>-3(3x)+5x

A:

Q: (2x3-x-5) - (6x3+x-x+1)

A:

Q: (x+1)(x2+2x-4)

A: To evaluate (x+1)(x2+2x-4)

Q: 5(-3x-2)-(x-3)=-4(4x+5)+13

A: The given equation is

Q: 4(x+1) = 3x-2 8

A: Consider the equation  4(x+1) = 3x-28

Q: Find (f 9)and indicate the domain of the quotient F(x)=5x*-20x and glx) = 3x²-1ax. %3D

A:

Q: 2x + 3x(x)-x+6x

A:

Q: 4x) = -2x* + 8x - 10 0 f-4)

A: To find value of a function at a point x= -4 , replace x by -4.

Q: g(9)=-2x2+x+13

A: Given: g(9)=-2x2+x+13       To determine: g(9)

Q: x2 + 8x + 15 = (x + 3)(x 4. x2 – 9x + 20 = (x – 5)(x.

A: Givenx2+8x+15we can rewrite 8x as sum of 3x and 5xx2+5x+3x+15taking common parts…

Q: (2x² - 4x+6)-(x² - 5)

A:

Q: Explain why f(x) = 5x + 7x + x + 9 has no real zeros.

A: Consider the given: Descartes' Rules of Signs:Let fx be a polynomial with real coefficients and a…

Q: x2 -x + 14 dx (x +2)(x² + 4)

A: We  have to find solution

Q: -8x² +13x+3 x(x+1)

A: We have to solve

Q: 4(x+4)=3+4x

A: The given equation is

Q: Find zeros

A: Answer Explanation:Step 1: Step 2: Step 3: Step 4:

Q: -2(x+5)-1/2(-10x-4)

A:

Q: 2(x + 3) - 7(2 - 3x)

A:

Q: Evaluate x2 - 2x for the value of x satisfying 4(x + 2) + -16 = 3x - 5(4 - x).

A: 4x+2+-16=3x-54-x⇒4x+8-16=3x-20+5x⇒4x-8=8x-20⇒8x-4x=20-8⇒4x=12⇒x=3

Q: 14 2x+ 3(x-2) 3(x-2)

A: