### Video Transcript

Determine the partial fraction
decomposition of π₯ squared plus π₯ plus one over π₯ multiplied by π₯ minus three
multiplied by π₯ plus one squared.

In order to answer this question,
we need to recall how we write an expression of the form ππ₯ squared plus ππ₯ plus
π over π₯ multiplied by π₯ minus π multiplied by π₯ minus π squared in partial
fractions. This can be written as four terms,
π΄ over π₯ plus π΅ over π₯ minus π plus πΆ over π₯ minus π plus π· over π₯ minus
π squared. We can then multiply each of the
five terms by the denominator on the left-hand side, π₯ multiplied by π₯ minus π
multiplied by π₯ minus π squared. The left-hand side becomes ππ₯
squared plus ππ₯ plus π. In the first term on the right-hand
side, the π₯s cancel. So weβre left with π΄ multiplied by
π₯ minus π multiplied by π₯ minus π squared.

In the second term, π₯ minus π
cancels. So weβre left with π΅π₯ multiplied
by π₯ minus π squared. In the third term, we can cancel
one of the π₯ minus πs. This leaves us with πΆπ₯ multiplied
by π₯ minus π multiplied by π₯ minus π. The final term becomes π·π₯
multiplied by π₯ minus π as the π₯ minus π squared cancels. In this question, the two brackets
or parentheses are π₯ minus three and π₯ plus one squared. This gives us π₯ squared plus π₯
plus one is equal to π΄ multiplied by π₯ minus three multiplied by π₯ plus one
squared plus π΅π₯ multiplied by π₯ plus one squared plus πΆπ₯ multiplied by π₯ minus
three multiplied by π₯ plus one plus π·π₯ multiplied by π₯ minus three.

There are two traditional methods
for calculating the values of π΄, π΅, πΆ, and π· from this point. One is by comparing
coefficients. However, in this case, as there are
so many terms, it will be easier to solve by substitution. The first two values that we will
choose, three and negative one, are obvious, as these are the numbers that would
make the parentheses zero. Negative one plus one is zero, and
three minus three is equal to zero. The Third value we will choose is
zero, as this would eliminate the second, third, and fourth term on the right-hand
side. We could choose any value for our
fourth substitution. In this case, weβll choose one, as
this will be easy to substitute.

Letβs begin by substituting π₯
equals three. On the left-hand side, we have
three squared plus three plus one. On the right-hand side, the first,
third, and fourth terms will become zero, as three minus three is zero and
multiplying any number by zero is still zero. We are, therefore, left with π΅
multiplied by three multiplied by three plus one squared. Three squared is equal to nine. Three plus one is equal to
four. Squaring this gives us 16 and
multiplying by three gives us 48. Therefore, the right-hand side
becomes 48π΅. Adding nine, three, and one gives
us 13. So 13 is equal to 48π΅. Dividing both sides by 48 gives us
a value of π΅ of 13 over 48.

Substituting in π₯ equals negative
one gives us negative one squared plus negative one plus one on the left-hand
side. On the right-hand side, the first
three terms will cancel as negative one plus one is zero. The final term becomes π·
multiplied by negative one multiplied by negative one minus three. Squaring negative one gives us
positive one. And then adding negative one and
adding one leaves us with one. The right-hand side simplifies to
four π·. Dividing both sides of this
equation by four gives us a value of π· equal to one-quarter.

Our next step is to substitute π₯
equal zero. The left-hand side becomes one. As zero squared plus zero plus one
is equal to one. The second, third, and fourth terms
all become zero. This leaves us with π΄ multiplied
by zero minus three multiplied by zero plus one squared on the right-hand side. Zero minus three is negative
three. Multiplying this by one and then by
π΄ gives us negative three π΄. Dividing both sides by negative
three gives us a value for π΄ equal to negative one-third.

Our final substitution is π₯ equals
one. On the left-hand side, we have
three, as one squared plus one plus one is equal to three. The first term on the right becomes
negative eight π΄. The second term is four π΅. The third term is negative four
πΆ. And the final term is negative two
π·. We can now substitute in the values
we have already calculated for π΄, π΅, and π·. Negative eight multiplied by
negative one-third is equal to eight-thirds or eight over three. Four multiplied by 13 over 48 is 13
over 12. Negative two multiplied by
one-quarter is negative one-half. The equation simplifies to three is
equal to eight-thirds plus 13 over 12 minus four πΆ minus one-half.

Our next step is to group the
constants on the right-hand side, eight-thirds plus 13 over 12 minus one-half. This is equal to 13 over four or 13
quarters. We can then subtract this from both
sides of the equation, giving us negative one-quarter is equal to negative four
πΆ. If negative one-quarter is equal to
negative four πΆ, we can divide both sides by negative four. This gives us a value of πΆ equal
to one 16th. We now have values for π΄, π΅, πΆ,
and π· of negative one-third, 13 over 48, one 16th, and one-quarter.

We can now write the initial
expression as four partial fractions. The first term is negative one over
three π₯. The second term is 13 over 48
multiplied by π₯ minus three. The third term is one over 16
multiplied by π₯ plus one. And the fourth and final term is
one over four multiplied by π₯ plus one squared. This is the partial fraction
decomposition of π₯ squared plus π₯ plus one over π₯ multiplied by π₯ minus three
multiplied by π₯ plus one squared.