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1. Which equation is best represented by the graph?
a. y= (x+3)(x+1)(x-1)
b. y= (x-3)(x+1)(x-1)
c. y= -(x-3)(x+1)(x-1)
d. y= (3-x)(x+1)(x-1)

2. Shown below is the graph of y=x^3-3x^2-6x+8. What are the apparent zeros of the function graphed above?

3. What is a polynomial function in a standard form with zeros 1,2,3 and -3?
a. x^4-3x^3+7x^2-27-18
b. x^4+3x^3-7x^2-27-18
c. x^4-3x^3-7x^2+27-18
d. x^4+3x^3+7x^2-27-18

4. What are the zeros of the function? What are their multiplicities?
F(x)= 5x^3-5x^2-30x
a. The num.. 3,-2, and 0 are zeros of multiplicity 1.
b. The num.. 3,-2, and 0 are zeros of multiplicity 2.
c. The num..-3,2, and 0 are zeros of multiplicity 1.
d. The num..-3,2, and 0 are zeros of multiplicity 2.

5. What are the real and complex solutions of the polynomial equation?
x^3-64=0(The / in the equations below aren't fractions, just so you know!)
a. 4,-1+2i/3,-1+2i/3
b. 4,1+2i/3,1+2i/3
c. 4,-2+2i/3,-2+2i/3
d. 4,+2+2i/3,+2+2i/3

6. What is the equivalent to the following expression?
a. 3p^2+8pq-3q^2
b. 4p^2+8pq-q^2
c. 4p^4+8pq-3q^4
d. 4p^2+8pq-3q^2

7. Divide -3x^3-4x^2+4x+3 by x-2.
a. -3x^2+2x+24
b. -3x^2-10x-16, R-29
c. -3x^2-10x-16
d. -3x^2+2x+24, R 35

8. Find the roots of the polynomial equation. X^3-4x^2+x+26=0
a. 3(+-)2i,-2
b. -3(+-)2i,-2
c. 3(+-)2i, 2
d. -3(+-)2i, 2

9. Which correctly describes the roots of the following cubic equation?
a. One real rt., two complex rts.
b. Tow real rts. and one complex rt.
c. Three real rts. two of which are equal in value
d. Three real rts, each with a different value

10. One zero of f(x)=x^3-2x^2-5x+6 is 1. What are other zeros of the function?
a. 3 and -2
b. -3 and 2
c. 1 and 2
d. 1 and -3

What of the following quartic functions has x= -1 and x= -3 as its only two real zeros?
a. x^4-4x^3-4x^2-4x-3
b. x^4+4x^3+4x^2+4x+3
c. x^4+4x^3+3x^2+4x-4
d. x^4+4x^3+4x^2+4x+3

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  1. sorry, no graphs on this web site.
    interesting problems.
    Did you try to do any of them?
    What are your answer choices?

    I'll do one to get you stated.
    #9 x^3+6x^2+11x+6
    by Descartes' Rule of Signs, there are 0 or 2 positive roots, and one or three negative root.
    by the Rational Root Theorem, the negative root must be -1, -2, -3 or -6
    A little checking show that x = -1 is one root, giving
    (x+1)(x^2 + 5x + 6) = (x+1)(x+2)(x+3)

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