Video Transcript
Consider π of π₯ is equal to π₯ to
the power of four plus three π₯ cubed minus five π₯ squared minus three π₯ plus
four. Write π of π₯ as the product of
linear and irreducible quadratic factors. List all zeros of π of π₯.
Our first step here is to try and
identify some roots of the quartic equation. The equation is a quartic as the
highest power is four. Therefore, it has a degree of
four. We can work out some roots of π of
π₯ by substituting in values of π₯. If our answer is equal to zero,
then the number weβve substituted in is a root.
We will start by considering π of
one, substituting π₯ equals one into the equation. This gives us one to the power of
four plus three multiplied by one cubed minus five multiplied by one squared minus
three multiplied by one plus four. One to the power of four is equal
to one. One cubed is equal to one. Multiplying this by three gives us
three. One squared is also equal to
one. And multiplying this by negative
five gives us negative five. Negative three multiplied by one is
equal to negative three.
Finally, we drop down the constant,
four. Adding the three positive terms β
one, three, and four β gives us eight. And adding the two negative terms β
negative five and negative three β gives us negative eight. As eight minus eight is equal to
zero, we can say that π of one is equal to zero. As one is a root, π₯ minus one is a
factor of π of π₯.
If this was a cubic equation, we
would then divide our cubic function by π₯ minus one. However, as this is a quartic, we
will repeat this process to find a second root of π of π₯. Weβll now consider π of negative
one, substituting in π₯ equals negative one. This gives us negative one to the
power of four plus three multiplied by negative one cubed minus five multiplied by
negative one squared minus three multiplied by negative one plus four. Negative one to the power of four
is equal to one. Negative one cubed is equal to
negative one. Multiplying this by three gives us
negative three. Negative one squared is equal to
positive one. Multiplying this by negative five
gives us negative five. Negative three multiplied by
negative one is equal to three.
Once again, we drop down the
constant, four. We can see once again that the
positive numbers sum to eight. The two negative numbers sum to
negative eight. Eight minus eight is equal to
zero. As π of minus one is equal to
zero, we can say that π₯ plus one is a factor. As π₯ minus one and π₯ plus one are
both factors of π of π₯, their product will also be a factor of π of π₯. We can expand or multiply out these
two parentheses using the FOIL method.
Multiplying the first terms gives
us π₯ squared. Multiplying the outside terms gives
us positive π₯. The inside terms multiply to give
negative π₯. And finally, the last terms
multiply to give us negative one. π₯ minus π₯ is equal to zero. Therefore, π₯ minus one multiplied
by π₯ plus one is equal to π₯ squared minus one.
We can now divide our quartic
function π of π₯ by π₯ squared minus one. We can do this following the
methods of division, multiplication, and then subtraction. Firstly, we divide π₯ to the power
of four by π₯ squared. This is equal to π₯ squared. We then multiply π₯ squared by π₯
squared minus one. This is equal to π₯ to the power of
four minus π₯ squared. We then subtract our two lines. π₯ to the four minus π₯ to the four
is equal to zero. Three π₯ cubed minus zero is three
π₯ cubed. And negative five π₯ squared minus
negative π₯ squared is negative four π₯ squared.
We drop down the next term in our
initial expression and repeat the process. Three π₯ cubed divided by π₯
squared is equal to three π₯. Multiplying three π₯ by π₯ squared
minus one gives us three π₯ cubed minus three π₯. We then subtract the two lines once
again. This gives us negative four π₯
squared, as three π₯ cubed minus three π₯ cubed is zero and negative three π₯ minus
negative three π₯ is also zero.
We drop down the positive four term
and repeat the process a third time. Negative four π₯ squared divided by
π₯ squared is equal to negative four. Multiplying negative four by π₯
squared minus one gives us negative four π₯ squared plus four. Subtracting these two lines gives
us an answer of zero, which means there is no remainder.
We will now clear some space by
removing our initial working. Our expression π of π₯, π₯ to the
power of four plus three π₯ cubed minus five π₯ squared minus three π₯ plus four, is
equal to π₯ squared minus one multiplied by π₯ squared plus three π₯ minus four. We already know that π₯ squared
minus one factorizes to π₯ minus one multiplied by π₯ plus one. Factorizing the second quadratic,
π₯ squared plus three π₯ minus four, gives us π₯ plus four multiplied by π₯ minus
one. Weβve now written π of π₯ as the
product of four linear factors.
We notice at this stage though that
two of the factors are identical. Theyβre π₯ minus one. This means that we can rewrite π
of π₯ as π₯ plus one multiplied by π₯ plus four multiplied by π₯ minus one all
squared. π of π₯ written as the product of
linear and irreducible quadratic factors is π₯ plus one multiplied by π₯ plus four
multiplied by π₯ minus one all squared.
The second part of our question
asked us to list all zeros of π of π₯. This is another way of saying write
all the roots of π of π₯. In our initial working, we
established that π₯ equals minus one and π₯ equals positive one were roots of π of
π₯. The third possible root comes from
the factor π₯ plus four and gives us an answer of π₯ equals negative four. This means that there are three
zeros of π of π₯. They are negative one, negative
four, and positive one.
