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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 #23 : Polynomial Functions

Zeros of a quadratic.

Find the zeros of

$\displaystyle f(x)=x^2-5x-36$.

Possible Answers:

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

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

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

$x=\{45.2,0.2\}$ $\displaystyle x=\{45.2,0.2\}$

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

Correct answer:

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

Explanation:

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

$\displaystyle (x-9)(x+4)$.

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

x-9=0 $\displaystyle x-9=0$,

so x=9. $\displaystyle x=9.$

And

x+4=0 $\displaystyle x+4=0$,

so x=-4 $\displaystyle x=-4$.

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Example Question #101 : Pre Calculus

Find the zeros of the following polynomial:

$\displaystyle x^2-12x+36$

Possible Answers:

6,3 $\displaystyle 6,3$

$\displaystyle 6$

$\displaystyle -6$

$\displaystyle 3$

$\displaystyle 5$

Correct answer:

$\displaystyle 6$

Explanation:

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

$\displaystyle x^2-12x+36=0$

$\displaystyle (x-6)^2=0$

x=6 $\displaystyle x=6$

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Example Question #102 : Pre Calculus

Simplify the following expression:

$\displaystyle x^4y^3z^2*x^2y^3z^4$

Possible Answers:

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

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

x^2z^2 $\displaystyle x^2z^2$

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

$\displaystyle x^6y^6z^6$

Correct answer:

$\displaystyle x^6y^6z^6$

Explanation:

When multiplying polynomials, add their exponents.

x:4+2=6 $\displaystyle x:4+2=6$

y:3+3=6 $\displaystyle y:3+3=6$

z:2+4=6 $\displaystyle z:2+4=6$

$\displaystyle x^6y^6z^6$

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Example Question #103 : Pre Calculus

Simplify this polynomial:

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

Possible Answers:

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

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

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

$\displaystyle x^6y^6z^6$

Correct answer:

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

Explanation:

When dividing polynomials, you must subtract corresponding exponents.

x:4-2=2 $\displaystyle x:4-2=2$

y:3-3=0 $\displaystyle y:3-3=0$

$\displaystyle z:2-4=-2$

Thus, our answer is

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

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Example Question #104 : Pre Calculus

Simplify the following:

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

Possible Answers:

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

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

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

$\frac{x}{y}$ $\displaystyle \frac{x}{y}$

Correct answer:

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

Explanation:

When dividing, we must subtract the exponents.

$\displaystyle x:4-(-3)=7$

$\displaystyle y=-1-2=-3$

z=2-2=0 $\displaystyle z=2-2=0$

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

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Example Question #105 : Pre Calculus

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

Possible Answers:

$\displaystyle x^3+6x^2-11x+6$

$\displaystyle x^3-3x^2+2x-1$

$\displaystyle x^3+3x^2-2x+1$

$\displaystyle x^3-6x^2+11x-6$

Correct answer:

$\displaystyle x^3-6x^2+11x-6$

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:

$\displaystyle (x-3)(x-2)(x-1)=0$

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

$\displaystyle (x^2-5x+6)(x-1)$

$\displaystyle x^3-5x^2+6x-x^2+5x-6$

$\displaystyle x^3-6x^2+11x-6$

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Example Question #106 : Pre Calculus

Solve for x:

$\displaystyle x(x-2)=3$

Possible Answers:

2,-3 $\displaystyle 2,-3$

3,-2 $\displaystyle 3,-2$

-3,1 $\displaystyle -3,1$

3,-1 $\displaystyle 3,-1$

Correct answer:

3,-1 $\displaystyle 3,-1$

Explanation:

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

$\displaystyle x(x-2)=3$

$\displaystyle x^2-2x=3$

$\displaystyle x^2-2x-3=0$

$\displaystyle (x-3)(x+1)=0$

x=3,-1 $\displaystyle x=3,-1$

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Example Question #1 : Solving Polynomial Equations

Solve the following for x:

$\displaystyle x^2(x+3)=4x$

Possible Answers:

-4,1 $\displaystyle -4,1$

-4,0,1 $\displaystyle -4,0,1$

4,-1 $\displaystyle 4,-1$

4,0,-1 $\displaystyle 4,0,-1$

Correct answer:

-4,0,1 $\displaystyle -4,0,1$

Explanation:

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

$\displaystyle x^2(x+3)=4x$

$\displaystyle x^3+3x^2-4x=0$

$\displaystyle x(x+4)(x-1)=0$

$\displaystyle x=0,-4,1$

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Example Question #4 : Solving Polynomial Equations

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

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

Possible Answers:

$\small \small h=2$ $\displaystyle \small \small h=2$

$\small \small h=-3$ $\displaystyle \small \small h=-3$

$\small \small h=3$ $\displaystyle \small \small h=3$

$\small h=30$ $\displaystyle \small h=30$

$\small \small h=90$ $\displaystyle \small \small h=90$

Correct answer:

$\small h=30$ $\displaystyle \small h=30$

Explanation:

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

$\small x+3 =0$ $\displaystyle \small x+3 =0$

$\small x=-3$ $\displaystyle \small x=-3$

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.

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

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

$\small 3h-90=0$ $\displaystyle \small 3h-90=0$

$\small 3h=90$ $\displaystyle \small 3h=90$

$\small h=30$ $\displaystyle \small h=30$

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Example Question #3 : Solving Polynomial Equations

Solve the following for x:

$\displaystyle x^2(x-6)=16x$

Possible Answers:

-2,0,8 $\displaystyle -2,0,8$

-2,8 $\displaystyle -2,8$

-1,0,16 $\displaystyle -1,0,16$

$\displaystyle 0$

-8,0,2 $\displaystyle -8,0,2$

Correct answer:

-2,0,8 $\displaystyle -2,0,8$

Explanation:

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

$\displaystyle x^2(x-6)=16x$

$\displaystyle x^3-6x^2-16x=0$

$\displaystyle x(x-8)(x+2)=0$

$\displaystyle x=-2,0,8$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #23 : Polynomial Functions

Example Question #101 : Pre Calculus

Example Question #102 : Pre Calculus

Example Question #103 : Pre Calculus

Example Question #104 : Pre Calculus

Example Question #105 : Pre Calculus

Example Question #106 : Pre Calculus

Example Question #1 : Solving Polynomial Equations

Example Question #4 : Solving Polynomial Equations

Example Question #3 : Solving Polynomial Equations

All Precalculus Resources

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