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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Polynomial Functions

Zeros of a quadratic.

Find the zeros of

f(x)=x^2-5x-36 .

Possible Answers:

$x=\{0,-9\}$ .

$x=\{-9,0\}$ .

$x=\{45.2,0.2\}$

$x=\{\frac{45}{2},-0\}$ .

$x=\{-4,9\}$ .

Correct answer:

$x=\{-4,9\}$ .

Explanation:

(-9)(4)=-36 , and -9+4=-5 , so the polynomial factors into

(x-9)(x+4) .

When the function is set to equal 0, either of the products that it factors into are 0. That is

x-9=0 ,

so x=9.

And

x+4=0 ,

so x=-4 .

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Example Question #101 : Pre Calculus

Find the zeros of the following polynomial:

x^2-12x+36

Possible Answers:

6,3

Correct answer:

Explanation:

To find the zeros, we must set it equal to zero, factor, and solve.

x^2-12x+36=0

(x-6)^2=0

x=6

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Example Question #1 : Algebra Of Functions

Simplify the following expression:

x^4y^3z^2*x^2y^3z^4

Possible Answers:

x^2z^2

$\frac{1}{x^6y^6z^6}$

$\frac{x^2}{z^2}$

$\frac{z^2}{x^2}$

x^6y^6z^6

Correct answer:

x^6y^6z^6

Explanation:

When multiplying polynomials, add their exponents.

x:4+2=6

y:3+3=6

z:2+4=6

x^6y^6z^6

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Example Question #2 : Algebra Of Functions

Simplify this polynomial:

$\frac{x^4y^3z^2}{x^2y^3z^4}$

Possible Answers:

$\frac{x^2}{z^2}$

$\frac{z^2}{x^2}$

$\frac{1}{x^6y^6z^6}$

x^6y^6z^6

Correct answer:

$\frac{x^2}{z^2}$

Explanation:

When dividing polynomials, you must subtract corresponding exponents.

x:4-2=2

y:3-3=0

z:2-4=-2

Thus, our answer is

$x^2y^0z^{-2}=\frac{x^2}{z^2}$ .

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Example Question #3 : Algebra Of Functions

Simplify the following:

$\frac{x^4y^{-1}z^2}{x^{-3}y^2z^2}$

Possible Answers:

$\frac{x}{y}$

$\frac{x^7z^4}{y^2}$

$\frac{x^7}{y^3}$

$\frac{x}{y^3}$

Correct answer:

$\frac{x^7}{y^3}$

Explanation:

When dividing, we must subtract the exponents.

x:4-(-3)=7

y=-1-2=-3

z=2-2=0

Thus, our answer is $\frac{x^7}{y^3}$ .

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Example Question #1 : Solving Polynomial Equations

A polynomial with leading term x^3 has roots 1,2, and 3. What is the polynomial?

Possible Answers:

x^3-6x^2+11x-6

x^3-3x^2+2x-1

x^3+6x^2-11x+6

x^3+3x^2-2x+1

Correct answer:

x^3-6x^2+11x-6

Explanation:

Given the roots of a polynomial, the problem can be solved in reverse. For 3,2, and 1 to be roots, the following must be true:

(x-3)(x-2)(x-1)=0

Therefore, expand the left side of the equation to find the polynomial.

(x^2-5x+6)(x-1)

x^3-5x^2+6x-x^2+5x-6

x^3-6x^2+11x-6

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Example Question #1 : Solving Polynomial Equations

Solve for x:

x(x-2)=3

Possible Answers:

3,-2

3,-1

-3,1

2,-3

Correct answer:

3,-1

Explanation:

To solve, we must expand, factor, and solve.

x(x-2)=3

x^2-2x=3

x^2-2x-3=0

(x-3)(x+1)=0

x=3,-1

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Example Question #2 : Solving Polynomial Equations

Solve the following for x:

x^2(x+3)=4x

Possible Answers:

-4,0,1

4,0,-1

-4,1

4,-1

Correct answer:

-4,0,1

Explanation:

To solve, we must expand, factor, and solve for x.

x^2(x+3)=4x

x^3+3x^2-4x=0

x(x+4)(x-1)=0

x=0,-4,1

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Example Question #4 : Solving Polynomial Equations

Given the polynomial $\small 4x^{3}+3x^{2}-hx-9$ , determine the value of $\small h$

given that the polynomial is exactly divisable by $\small x+3$

Possible Answers:

$\small \small h=2$

$\small \small h=-3$

$\small \small h=3$

$\small h=30$

$\small \small h=90$

Correct answer:

$\small h=30$

Explanation:

First determine the value of x by setting $\small x+3$ equal to 0.

$\small x+3 =0$

$\small x=-3$

Next, plug in the value of x determined above into the polynomial. The polynomial is to be set equal to 0. Solve for the value of h.

$\small \small \small 4(-3)^{3}+3(-3))^{2}-(-3)h-9=0$

$\small -108+27+3h-9=0$

$\small 3h-90=0$

$\small 3h=90$

$\small h=30$

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Example Question #4 : Solving Polynomial Equations

Solve the following for x:

x^2(x-6)=16x

Possible Answers:

-2,8

-8,0,2

-1,0,16

-2,0,8

Correct answer:

-2,0,8

Explanation:

To solve, we must expand, factor, and solve for x.

x^2(x-6)=16x

x^3-6x^2-16x=0

x(x-8)(x+2)=0

x=-2,0,8

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Polynomial Functions

Example Question #101 : Pre Calculus

Example Question #1 : Algebra Of Functions

Example Question #2 : Algebra Of Functions

Example Question #3 : Algebra Of Functions

Example Question #1 : Solving Polynomial Equations

Example Question #1 : Solving Polynomial Equations

Example Question #2 : Solving Polynomial Equations

Example Question #4 : Solving Polynomial Equations

Example Question #4 : Solving Polynomial Equations

All Precalculus Resources

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