5.4- solve higher-order polynomials equations and inequalities in one and two variables

a³+b³=

(a+b)(a²+b²-ab)

a³-b³=

(a-b)(a²+b²+ab)

x³+y³+z³-3xyz=

(x+y+z)(x²+y²+z²-xy-yz-zx)

what is x²-3x-10 ÷ x+2?

(x-5)(x+2)/x+2 cancel out x+2 therefore, x-5 is a common factor

what polynomial must be added to 3x²+4x+7 to obtain a sum if 0?

-3x²-4x-1

3x²-3-7x-6x²+5, combine like terms...?

-3x²-7x+2

divide: (-30x²-12x+18)÷6?

-5x²-2x+3 6|-30x²-12x+18

multiply: (5y-7)(-y)..?

-5y²+7y

the expression (-6+i²) is equivalent to which of the following expressions?

1. (6-i)²=(-6+i)(-6+i) 2. 36-6i-6i+i² 3. 36-12i+i² 4. 36-12i-1 35-12i

f(x)=(2x-1)(x-5), find the zeros...?

1. Set the function equal to 0. 2. 2x-1=0 and x-5=0 {½, -5}

f(x)=(x-3)(3x+1)(x+1), find the zeros...?

1. Set the function equal to 0. 2. x-3=0, 3x+1=0, and x+1=0 {3,-⅓, -1}

f(x)=x(2x-1)(x-1)(x+1), find the zeros...?

1. Set the function equal to 0. 2. x=0, 2x-1=0, x-1=0, and x+1=0 {0, ½, 1, -1}

f(x)=x(5x-2)(x²+1), find the zeros...?

1. Set the function equal to 0. 2. x=0, 5x-2=0, x²+1=0, and x+1=0 {0,⅖, ±1}

f(x)=x(x-2)(x-2)(3x²-4), find the zeros...?

1. Set the function equal to 0. 2. x=0, x+2=0, x-2=0, and 3x²-4=0 {0 -2, 2, 3x²=4} 3. Add 4 to both sides of the equation, x²= √(4/3) 4. take the square root of both sides of the equation to eliminate the exponent on the left side, x=±√(4/3) 5. Simplify the right side of the equation, rewrite √(4/3) as √4/√3, x=±√4/√3 6. Simplify the numerator. x=±2/√3 7. Multiply 2/√3 by √3/√3, x=±2/√3×√3/√3 = 2√3/3, -2√3/3

f(x)= x⁴-x²-12, find all the zeros steps...?

1. Set x⁴-x²-12 equal to 0. 2. Solve for x. 3. Substitute u=x² into the equation. This will make the quadratic formula easy to use, u²-u-12=0...u=x² 4. Factor u²-u-12 using the AC method. (u-4)(u+3)=0. The final solution is all the values that make (u-4)(u+3)=0 true u=4, -3. 5. Substitute the real value of u=x back into the solved equation. x²=4...(x²)¹=-3 6. Solve the equation for x=2, -2, x=±√3i

f(x)= x⁴-x²-30, find all the zeros steps...?

1. Set x⁴-x²-30 equal to 0. 2. Solve for x. 3. Substitute u=x² into the equation. This will make the quadratic formula easy to use, u²-u-30=0...u=x² 4. Factor u²-u-30 using the AC method. (u-6)(u-5)=0. The final solution is all the values that make (u-6)(u+5)=0 true u=6, -5. 5. Substitute the real value of u=x² back into the solved equation. x²=6...(x²)¹=-5 6. Solve the equation for x., x=√6, -√6; (x²)¹=-5...x=i√5, -i√5

f(x)= x³+8, find all the zeros steps...?

1. Subtract 8 from both sides of the equation. 2. Factor the left side of the equation. 3. Rewrite 8 as 2³, x³+2³=0 4. Since both terms are perfect cubes factor using the sum of cubes formula, a³+b³=(a+b)(a²-ab+b²) where a=x and b=2, (x+2)(x²-x×2+2²)=0 5. Simplify. (x+2)(x²-2x+4) 6. If any individual Factor on the left side of the equation is equal to zero the entire spection will be equal to 0., x+2=0, x²-2x+4=0 7. Set x+2 and x²-2x+4 equal to 0 and solve for x. 8. Use the quadratic formula to find the solutions.

steps to find roots and the equation for 3, 1, -2. -4...?

1. Which of the points where the graph intercepts with the x axis (y=0) 2. The root at x=3 was found by solving for x when x-(3)=y and y=0. The factor is x-3.....,do the same with 2, -2... 3. Combine all the factors into a single equation, y=(x-3)(x-2)(x+2) 4. Multiply all the factors to simplify the equation, y=(x-3)(x-2)(x+2) 5. Expand (x-3)(x-2) using the FOIL Method... y=x³-3x²-4x+12

what value of k makes q⁴¹=q^4k×q⁵ true?

1. q⁴¹=q^4k×q⁵ 2. q⁴¹=q^4k+5 3. The bases are equal, so equate the powers. 41=4k+5 41-5=4k 36=4k K=9

(3x²-4x+8) and (-x²-2x-8) combine like terms...?

2x²-6x+0

Which of the following equation should be solved in determining the roots of 2x³-8x²-8x+32=0

2x³-8x²-8x+32=0 2x²(x-4)-8(x-4)=0 2(x-2)(x+2)(x-4)=0 this question requires the examinee to solve higher order polynomial equations and inequalities.

[(3x²+5xy)-(6x²-4xy)]÷4x

3x²+5xy-6x²+4xy =(-3x²+9xy)/4x =-¾x+9/4y

divide: 12x²÷3...?

4x²

The degree of a polynomial function f(x) is odd. This implies that the graphs of f(x) has:

5.4- Competency 5 this question requires the examinee to analyze the relationship between a higher order polynomial function and it's graph. Since every polynomial function with real coefficients and positive degree n can be uniquely factored into a product of linear factors and irreducible quadratic factors and/or irreducible quadratic factors, a polynomial of odd degree Has at least 1 X intercept.

Subtract (5r²-4)- (8r²+7r+8)

5r²-4- (8r²+7r+8) -3r²-7r-12

divide: 9x²÷-3x...?

9x²/-3x= -3x

(2a³b^-2)(-4a²b⁴), simplify...?

= -8a³⁺²b²⁺⁴ (product of powers) = -8a⁵b⁶ (simplify)

Polynomial

A monomial or the sum of monomials

Factor Theorem for Polynomials

For any polynomial function, p(x), if the remainder after division by the factor (x - c) equals 0, then (x - c) is a factor of the polynomial. also if we divide a polynomial by x - c we obtain a result of the form: f(x)=(x-c)q(x)+f(c)

arithmetic operations polynomial

addition, subtraction, multiplication, division of a polynomial

why are properties of exponents useful in in astronomy?

astronomy deals with very large numbers that are sometimes difficult to work with because they contain so many digits. Properties of exponents make very large or very small numbers more manageable. As long as you know how far away a planet is from a light source you can divide that Distance by the speed of light to obtain how long it will take light to reach the planet.

(a+b)³=

a³+b³+3ab(a+b)

(a+b+c)³=

a³+b³+c³+3(a+b)(b+c)(c+a)

(a-b)³=

a³-b³-3ab(a-b)

find a factor of f(x)=x²-3x-4...?

f(0)=0-3(0)²-4 f(0)=4 f(x)=x²-3(4)-4=16-12-4=0...so x-4 must be a factor of x²-3x-4

find a factor of f(x)=x⁴-2x²+1...?

f(1)=1⁴-2(1)²+1 f(1)=0 f(x)=x⁴-2x²+1...has a factor of x-1

why is x-1 not a factor of 2x³-x²+18?

f(1)=2×1³-1²-21×1+18=2-1-21+18=-2 therefore f(1)≠0

is a+9 a factor of 3a²+21a-51, yes or no?

no, a+9 is not a factor

is a+3 a factor of 9a²+21a-18, yes or no?

yes, a+3 is a factor