Roots of Polynomials - GRE Subject Test: Math

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Question

What are the roots of the polynomial: ?

Answer

Step 1: Find factors of 44:

Step 2: Find which pair of factors can give me the middle number. We will choose .

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:

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 and .

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Question

Solve for :

Answer

Step 1: Factor by pairs:

Step 2: Re-write the factorization:

Step 3: Solve for x:

$\ x^2=-1;x=4 \\sqrt{x^2}=\pm\sqrt{-1}; x=4 \x=\pm i;x=4$

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Question

Find :

Answer

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

We will choose .

Step 2: Factor using the numbers we chose:

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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Question

What are the roots of ?

Answer

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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Question

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

Answer

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:

First group the terms into two binomials:

Then take out the greatest common factor from each group:

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

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

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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Question

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

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

Answer

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

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

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

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

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

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

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

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Question

Find all of the roots for the polynomial below:

Answer

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

Group into two binomials:

Take out the greatest common factor from each binomial:

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

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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Question

Find all of the roots for the polynomial below:

Answer

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

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

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

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

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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