Question 193068
# 1


To factor {{{6x^3 + 31x^2 + 4x - 5}}}, we can use synthetic division



First, let's find our test zero:


{{{x+5=0}}} Set the given factor {{{x+5}}} equal to zero


{{{x=-5}}} Solve for x.


so our test zero is -5



Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of {{{6x^3 + 31x^2 + 4x - 5}}} to the right of the test zero.<TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)

<TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -5 by 6 and place the product (which is -30)  right underneath the second  coefficient (which is 31)

    <TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-30</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -30 and 31 to get 1. Place the sum right underneath -30.

    <TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-30</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>1</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -5 by 1 and place the product (which is -5)  right underneath the third  coefficient (which is 4)

    <TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-30</TD><TD>-5</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>1</TD><TD></TD><TD></TD></TR></TABLE>

    Add -5 and 4 to get -1. Place the sum right underneath -5.

    <TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-30</TD><TD>-5</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>1</TD><TD>-1</TD><TD></TD></TR></TABLE>

    Multiply -5 by -1 and place the product (which is 5)  right underneath the fourth  coefficient (which is -5)

    <TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-30</TD><TD>-5</TD><TD>5</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>1</TD><TD>-1</TD><TD></TD></TR></TABLE>

    Add 5 and -5 to get 0. Place the sum right underneath 5.

    <TABLE cellpadding=10><TR><TD>-5</TD><TD>|</TD><TD>6</TD><TD>31</TD><TD>4</TD><TD>-5</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-30</TD><TD>-5</TD><TD>5</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>6</TD><TD>1</TD><TD>-1</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+5}}} is a factor of  {{{6x^3 + 31x^2 + 4x - 5}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (6,1,-1) form the quotient


{{{6x^2 + x - 1}}}



So  {{{6x^3 + 31x^2 + 4x - 5}}} factors to {{{(x+5)(6x^2 + x - 1)}}}



In other words, {{{6x^3 + 31x^2 + 4x - 5=(x+5)(6x^2 + x - 1)}}}



I'll let you continue the factorization....



<hr>



# 2



First lets find our test zero:


{{{x+1=0}}} Set the denominator {{{x+1}}} equal to zero


{{{x=-1}}} Solve for x.


so our test zero is -1



Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of {{{x^4 -2x3 + x^2 -4}}} to the right of the test zero.(note: remember if a polynomial goes from {{{1x^2}}} to {{{-4x^0}}} there is a zero coefficient for {{{x^1}}}. This is simply because {{{x^4 - 2x^3 + x^2 - 4}}} really looks like {{{1x^4+-2x^3+1x^2+0x^1+-4x^0}}}<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by 1 and place the product (which is -1)  right underneath the second  coefficient (which is -2)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -1 and -2 to get -3. Place the sum right underneath -1.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by -3 and place the product (which is 3)  right underneath the third  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>3</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 3 and 1 to get 4. Place the sum right underneath 3.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>3</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by 4 and place the product (which is -4)  right underneath the fourth  coefficient (which is 0)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>3</TD><TD>-4</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Add -4 and 0 to get -4. Place the sum right underneath -4.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>3</TD><TD>-4</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD><TD></TD></TR></TABLE>

    Multiply -1 by -4 and place the product (which is 4)  right underneath the fifth  coefficient (which is -4)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>3</TD><TD>-4</TD><TD>4</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD><TD></TD></TR></TABLE>

    Add 4 and -4 to get 0. Place the sum right underneath 4.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>1</TD><TD>0</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>3</TD><TD>-4</TD><TD>4</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+1}}} is a factor of  {{{x^4 - 2x^3 + x^2 - 4}}}


Now lets look at the bottom row of coefficients:


The first 4 coefficients (1,-3,4,-4) form the quotient


{{{x^3 - 3x^2 + 4x - 4}}}



So  {{{x^4 - 2x^3 + x^2 - 4}}} factors to {{{(x+1)(x^3 - 3x^2 + 4x - 4)}}}



In other words, {{{x^4 - 2x^3 + x^2 - 4=(x+1)(x^3 - 3x^2 + 4x - 4)}}}



Now let's use the factor {{{x-2}}} to factor {{{x^3 - 3x^2 + 4x - 4}}}



First lets find our test zero:


{{{x-2=0}}} Set the denominator {{{x-2}}} equal to zero


{{{x=2}}} Solve for x.


so our test zero is 2



Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of {{{x^3 - 3x^2 + 4x - 4}}} to the right of the test zero.<TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

<TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by 1 and place the product (which is 2)  right underneath the second  coefficient (which is -3)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 2 and -3 to get -1. Place the sum right underneath 2.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by -1 and place the product (which is -2)  right underneath the third  coefficient (which is 4)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>-2</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD></TR></TABLE>

    Add -2 and 4 to get 2. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>-2</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>2</TD><TD></TD></TR></TABLE>

    Multiply 2 by 2 and place the product (which is 4)  right underneath the fourth  coefficient (which is -4)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>-2</TD><TD>4</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>2</TD><TD></TD></TR></TABLE>

    Add 4 and -4 to get 0. Place the sum right underneath 4.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>1</TD><TD>-3</TD><TD>4</TD><TD>-4</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>-2</TD><TD>4</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>2</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x-2}}} is a factor of  {{{x^3 - 3x^2 + 4x - 4}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,-1,2) form the quotient


{{{x^2 - x + 2}}}



So {{{(x^3 - 3x^2 + 4x - 4)/(x-2)=x^2 - x + 2}}}



Basically  {{{x^3 - 3x^2 + 4x - 4}}} factors to {{{(x-2)(x^2 - x + 2)}}}



So {{{x^3 - 3x^2 + 4x - 4=(x-2)(x^2 - x + 2)}}}



This means that {{{x^4 - 2x^3 + x^2 - 4=(x+1)(x^3 - 3x^2 + 4x - 4)}}} then becomes



{{{x^4 - 2x^3 + x^2 - 4=(x+1)(x-2)(x^2 - x + 2)}}}



So all you have to do now is factor {{{x^2 - x + 2}}} (I'll let you do that)