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GRE Subject Test: Math : Classifying Algebraic Functions

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Example Question #1 : Roots Of Polynomials

What are the roots of the polynomial: f(x)=x^2-7x-44 ?

Possible Answers:

x=-11,x=-4

x=11,-4

x=-11, x=4

None of the Above

Correct answer:

x=11,-4

Explanation:

Step 1: Find factors of 44:

(1,44), (2,22), (4,11)

Step 2: Find which pair of factors can give me the middle number. We will choose (4,11) .

Step 3: Using and , we need to get . The only way to get is if I have and .

Step 4: Write the factored form of that trinomial:

(x-11)(x+4)

Step 5: To solve for x, you set each parentheses to :

$(x-11)=0\rightarrow x=11$
$(x+4)=0\rightarrow x=-4$

The solutions to this equation are x=11 and x=-4 .

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Example Question #1 : Algebra

Solve for :

x^3-4x^2+x-4

Possible Answers:

$x=\pm1,-4$

x=1,4

x=-1,4

$x=\pm i,4$

Correct answer:

$x=\pm i,4$

Explanation:

Step 1: Factor by pairs:

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

x^2(x-4)+1(x-4)=0

Step 2: Re-write the factorization:

(x^2+1)(x-4)=0

Step 3: Solve for x:

(x^2+1)=0;(x-4)=0
$\\ x^2=-1;x=4 \\\sqrt{x^2}=\pm\sqrt{-1}; x=4 \\x=\pm i;x=4$

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Example Question #1 : Algebra

Find :

x^2+5x-6

Possible Answers:

No Solutions Exist

x=1

x=1,-6

x=6

Correct answer:

x=1,-6

Explanation:

Step 1: Find two numbers that multiply to and add to .

We will choose 6,-1 .

Step 2: Factor using the numbers we chose:

(x+6)(x-1)=0

Step 3: Solve each parentheses for each value of x..

$(x+6)=0 \implies x=-6$
$(x-1)=0\implies x=1$

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Example Question #1 : Algebra

$Find\ all\ of\ the\ roots\ for\ the\ below\ polynomial:$

x^3+5x^2-3x-15=0

Possible Answers:

$x= \sqrt{3}\ and\ -5$

$x=3\ and\ 15$

$x=\pm \sqrt{3}\ and\ -5$

$x=3\ and\ -15$

Correct answer:

$x=\pm \sqrt{3}\ and\ -5$

Explanation:

Based upon the fundamental theorem of algebra, we know that there must exist 3 roots for this polynomial based upon its' degree of 3.

To solve for the roots, we use factor by grouping:

x^3+5x^2-3x-15=0

First group the terms into two binomials:

(x^3+5x^2)(-3x-15)=0

Then take out the greatest common factor from each group:

x^2(x+5)-3(x+5)=0

Now we see that the leftover binomial is the greatest common factor itself:

$(x+5)\ is\ common\ to\ both\ terms\ so\ we\ can\ take\ it\ out:$

(x+5)(x^2-3)=0

We set each binomial equal to zero and solve:

$x+5=0\ so\ x=-5$

$x^2-3=0;\ so\ x^2=3\ which\ means:$

$x=\pm\sqrt{3}\ after\ taking\ the\ square\ root\ of\ both\ sides.$

$The\ 3\ roots\ are\ -5, {\sqrt{3}},\ and\ -\sqrt{3}$

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Example Question #1 : Polynomials

Find all of the roots for the polynomial below:

f(x)=x^3+21-7x^2-3x

Possible Answers:

$-21,\ 3\ and\ \frac{3}{7}$

$7,\ and\ \sqrt{3}$

$3\ and\ \frac{3}{7}$

$7,\ \sqrt{3}\ and\ -\sqrt{3}$

Correct answer:

$7,\ \sqrt{3}\ and\ -\sqrt{3}$

Explanation:

In order to find the roots for the polynomial we must first put it in Standard Form by decreasing exponent:

f(x)=x^3+21-7x^2-3x

x^3-7x^2-3x+21

Now we can use factor by grouping, we start by grouping the 4 terms into 2 binomials:

(x^3-7x^2)(-3x+21)

We now take the greatest common factor out of each binomial:

x^2(x-7)-3(x-7)

We can see that each term now has the same binomial as a common factor, so we simplify to get:

(x-7)(x^2-3)

To find all of the roots, we set each factor equal to zero and solve:

$x-7=0\ which\ gives\ us\ x=7$

$x^2-3=0\ which\ gives\ us\ x^2=3\ which\ then\ yields\ x=\pm \sqrt{3}$

$The\ roots\ are\ 7,\ \sqrt{3}\ and\ -\sqrt{3}$

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Example Question #6 : Roots Of Polynomials

$How\ many\ total\ roots\ will\ a\ polynomial\ with\ the\ below\ roots\ have?$

$x=-\sqrt{5};\ x=7; and\ x=3+2i$

Possible Answers:

Correct answer:

Explanation:

$If\ the\ complex\ root\ a+bi\ exists, then\ a-bi\ must\ also\ exist\ as\ a\ root.$

$If\ the\ irrational\ root\ a+\sqrt{b}\ exists,\ then\ the\ root\ a-\sqrt{b}\ must\ also\ exist.$

$Given\ that\ the\ above\ statements\ always\ hold\ true\ we\ know\ that:$

$If\ -\sqrt{5}\ exists\ as\ a\ root\, then\ +\sqrt{5}\ also\ exists\ as\ a\ root.$

$Also,\ if\ 3+2i\ exists\ as\ a\ root,\ then\ 3-2i\ must\ also\ exist\ as\ a\ root.$

$This\ gives\ us\ an\ additional\ 2\ roots\ above\ the\ 3\ roots\ we\ already\ know.$

$There\ will\ be\ a\ total\ of\ 5\ roots.$

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Example Question #1 : Polynomials

What are the roots of x^2-3x-10 ?

Possible Answers:

-5,2

5,-2

Correct answer:

5,-2

Explanation:

Step 1: Find two numbers that multiply to $\large -10$ and add to $\large -3$ ...

We will choose $\large -5, 2$

Check:

$\large (5)(-2)=-10$
$\large -5+2=-3$

We have the correct numbers...

Step 2: Factor the polynomial...

$\large x^2-3x-10=(x-5)(x+2)$

Step 3: Set the parentheses equal to zero to get the roots...

$\large (x-5)=0\implies x=5$

$\large (x+2)=0 \implies x=-2$

So, the roots are $\large \large 5, -2$ .

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Example Question #1 : Classifying Algebraic Functions

Find all of the roots for the polynomial below:

f(x)=x^3+5x^2-4x-20

Possible Answers:

$-5\ and\ 2$

$-5,\ 2\, and\ -2$

$-5,\ 4\ and\ -4$

$-4\ and\ 5$

Correct answer:

$-5,\ 2\, and\ -2$

Explanation:

In order to find the roots of the polynomial we must factor by grouping:

x^3+5x^2-4x-20

Group into two binomials:

(x^3+5x^2)(-4x-20)

Take out the greatest common factor from each binomial:

x^2(x+5)-4(x+5)

We can now see that each term has a common binomial factor:

(x+5)(x^2-4)

We set each factor equal to zero and solve to obtain the roots:

$x+5=0\ and\ x^2-4=0$

$x=-5\ and\ x^2=4$

$x=-5\ and\ \sqrt{x^2}=\sqrt{4}\ so\ x=\pm2$

$The\ roots\ are\ -5,\ 2\ and\ -2$

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Example Question #1 : Binomial Expansion

Expand: (x+2)^4 .

Possible Answers:

x^4+8x^3+25x^2+32x+16

x^4-8x^3+24x^2-32x+16

-x^4+8x^3-24x^2+32x-16

x^4+8x^3+24x^2+32x+16

Correct answer:

x^4+8x^3+24x^2+32x+16

Explanation:

Step 1: Evaluate (x+2)^2 .

(x+2)^2=x^2+4x+4

Step 2. Evaluate (x+2)^3

From the previous step, we already know what (x+2)^2 is.

(x+2)^3 is just multiplying by another (x+2)

(x^2+4x+4)(x+2)=x^3+2x^2+4x^2+8x+4x+8
=x^3+6x^2+12x+8

Step 3: Evaluate (x+2)^4 .

(x+2)^4=(x+2)^3(x+2)

(x^3+6x^2+12x+8)(x+2)
=x^4+2x^3+6x^3+12x^2+12x^2+24x+8x+16
=x^4+8x^3+24x^2+32x+16

The expansion of (x+2)^4 is

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Example Question #1 : Binomial Expansion

What is the expansion of (x+1)^6 ?

Possible Answers:

x^5+5x^4+10x^3+10x^2+5x+1

x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1

x^4+4x^3+6x^2+4x+1

x^6+6x^5+15x^4+20x^3+15x^2+6x+1

Correct answer:

x^6+6x^5+15x^4+20x^3+15x^2+6x+1

Explanation:

Solution:

We can look at Pascal's Triangle, which is a quick way to do Binomial Expansion. We read each row (across, left to right)

For the first row, we only have a constant.
For the second row, we get x+constant .
...
For the 7th row, we will start with an x^6 term and end with a constant.

Ptriangle

Step 1: We need to locate the 7th row of the triangle and write the numbers in that row out.

The 7th row is 1,6,15,20,15,6,1 .

Step 2: If we translate the 7th row into an equation, we get:

x^6+6x^5+15x^4+20x^3+15x^2+6x+1 . This is the solution.

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GRE Subject Test: Math : Classifying Algebraic Functions

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Roots Of Polynomials

Example Question #1 : Algebra

Example Question #1 : Algebra

Example Question #1 : Algebra

Example Question #1 : Polynomials

Example Question #6 : Roots Of Polynomials

Example Question #1 : Polynomials

Example Question #1 : Classifying Algebraic Functions

Example Question #1 : Binomial Expansion

Example Question #1 : Binomial Expansion

All GRE Subject Test: Math Resources

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