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HiSET: Math : Zeroes of polynomials

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Example Questions

Example Question #42 : Algebraic Concepts

The equation

$x^{2} + 34 = 17x$

has two distinct solutions. What is their sum?

Possible Answers:

Correct answer:

Explanation:

It is not necessary to actually find the solutions to a quadratic equation to determine the sum of its solutions.

First, get the equation in standard form $ax^{2}+ bx + c = 0$ by subtracting 17x from both sides:

$x^{2} + 34 = 17x$

$x^{2} + 34 -17x = 17x - 17x$

$x^{2} -17x + 34 = 0$

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, b= -17 , making the correct choice.

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Example Question #271 : Hi Set: High School Equivalency Test: Math

The graph of the polynomial function

$f(x)= x^{3}+x^{2}-5x+2$

has one and only one zero on the interval (0, 1) . On which subinterval is it located?

Possible Answers:

(0,6, 0.8)

(0, 0.2)

(0.4, 0.6)

(0.8, 1)

(0.2, 0.4)

Correct answer:

(0.4, 0.6)

Explanation:

The Intermediate Value Theorem (IVT) states that if the graph of a function is continuous on an interval [a, b] , and f(a) and f(b) differ in sign, then has a zero on (a,b) . Consequently, the way to answer this question is to determine the signs of on the endpoints of the subintervals - $\left \{ 0, 0.2, 0.4, 0.6, 0.8, 1 \right \}$ . We can do this by substituting each value for as follows:

$f(x)= x^{3}+x^{2}-5x+2$

$f(0)= 0^{3}+0^{2}-5(0)+2 = 0+0-0+2= 2$

$f(0.2)= 0.2^{3}+0.2^{2}-5(0.2)+2 = 0.008+0.04-1+2 = 1.048$

$f(0.4)= 0.4^{3}+0.4^{2}-5(0.4)+2= 0.064+ 0.16- 2+2 = 0.224$

$f(0.6)= 0.6^{3}+0.6^{2}-5(0.6)+2= 0.216+ 0.36- 3+2 = -0.424$

$f(0.8)= 0.8^{3}+0.8^{2}-5(0.8)+2= 0.512+ 0.64- 4+2 = -0.848$

$f(1)=1^{3}+1^{2}-5(1)+2 =1+1-5+2=-1$

assumes positive values for x= 0, 0.2, 0.4 and negative values for x = 0.6, 0.8, 1 . By the IVT, has a zero on (0.4, 0.6) .

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Example Question #43 : Algebraic Concepts

The graph of a function is shown below, with labels on the y-axis hidden.

Graph zeroes 2

Determine which of the following functions best fits the graph above.

Possible Answers:

$f(x)=x^2x^{-2}$

f(x)=-2x^2(x+1)(x-2)(x+2)

$f(x)=\frac{x+1}{x+2}$

f(x)=-2(x+1)(x-2)(x+2)

f(x)=(x+1)^3(x-2)^2

Correct answer:

f(x)=-2(x+1)(x-2)(x+2)

Explanation:

Use the zeroes of the graph to determine the matching function. Zeroes are values of x where f(x)=0 . In other words, they are points on the graph where the curve touches zero.

Visually, you can see that the curve crosses the x-axis when x=-2 , x=-1 , and x=2 . Therefore, you need to look for a function that will equal zero at these x values.

A function with a factor of (x+2) will equal zero when x=-2 , because the factor of (x+2) will equal zero. The matching factors for the other two zeroes, x=-1 and x=2 , are (x+1) and (x-2) , respectively.

The answer choice f(x)=-2x^2(x+1)(x-2)(x+2) has all of these factors, but it is not the answer because it has an additional zero that would be visible on the graph. Notice it has a factor of x^2 , which results in a zero at x=0 . This additional zero that isn't present in the graph indicates that this cannot be matching function.

f(x)=-2(x+1)(x-2)(x+2) is the answer because it has all of the required factors and, as a result, the required zeroes, while not having additional zeroes. Notice that the constant coefficient of negative 2 does not affect where the zeroes are.

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Example Question #2 : Use The Zeros To Construct A Rough Graph Of A Function

Graph zeroes

Which of the functions below best matches the graphed function?

Possible Answers:

f(x)= (3x+5)(x+2)

f(x)= x(3x-5)(x-2)

f(x)= x(3x-5)(x+2)

f(x)= x(x+2)^2

f(x)= x(3x+5)(x-2)

Correct answer:

f(x)= x(3x-5)(x+2)

Explanation:

First, look at the zeroes of the graph. Zeroes are where the function touches the x-axis (i.e. values of where f(x)=0 ).

The graph shows the function touching the x-axis when x=-2 , x=0 , and at a value in between 1.5 and 2.

Notice all of the possible answers are already factored. Therefore, look for one with a factor of (which will make f(x)=0 when x=0 ), a factor of (x+2) to make f(x)=0 when x=-2 , and a factor which will make f(x)=0 when is at a value between 1.5 and 2.

f(x)= x(3x-5)(x+2)

This function fills the criteria; it has an and an (x+2) factor. Additionally, the third factor, (3x-5) , will result in f(x)=0 when $x=\frac{5}{3}$ , which fits the image. It also does not have any extra zeroes that would contradict the graph.

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HiSET: Math : Zeroes of polynomials

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #42 : Algebraic Concepts

Example Question #271 : Hi Set: High School Equivalency Test: Math

Example Question #43 : Algebraic Concepts

Example Question #2 : Use The Zeros To Construct A Rough Graph Of A Function

All HiSET: Math Resources

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