# Video: CBSE Class X • Pack 1 • 2018 • Question 14

CBSE Class X • Pack 1 • 2018 • Question 14

07:56

### Video Transcript

Find all zeros of the polynomial two 𝑥 to the power of four minus nine 𝑥 cubed plus five 𝑥 squared plus three 𝑥 minus one if two of its zeros are two plus the square root of three and two minus the square root of three.

To begin this question, let’s set up some notation. We’ll call the polynomial given by our question the function 𝑓 of 𝑥. Now, the question asked us to find all the zeros of this polynomial. What this means is that we need to find all of the 𝑥-values that make our function 𝑓 of 𝑥 equal to zero.

A quick thing to look out for in this type of question is the highest power of 𝑥 in your polynomial. For our polynomial, the highest power of 𝑥 we have is 𝑥 to the four. This tells us that the polynomial will have up to four zeros or four roots that we need to look out for.

Now, our end goal will be to find some factorization for the given polynomial, which is the same or similar to the form shown here. Once we found this type of factorization, we will be able to say that if 𝑥 is equal to 𝑎, then we see that our first set of brackets is equal to 𝑎 minus 𝑎 which is zero.

Since anything times zero is also zero, we will therefore have found one of the points where 𝑓 of 𝑥 is equal to zero. By similar reasoning, if 𝑥 is any of the values 𝑎, 𝑏, 𝑐, or 𝑑, then 𝑓 of 𝑥 will be zero. And we will have found the zeros of the polynomial.

To recap what this example tells us, we know that if 𝑎 is a zero of our polynomial, then 𝑥 minus 𝑎 will be a factor of the polynomial. Now, the question has already given us two zeros to start things off. Two plus the square root of three and two minus the square root of three.

From this, we know that when 𝑓 of 𝑥 equals zero, 𝑥 equals two plus the square root of three and 𝑥 equals two minus the square root of three. We now know that these zeros can be used to find two of the factors of our polynomial. We have 𝑥 minus two plus the square root of three and 𝑥 minus two minus the square root of three.

Now, we know that these are both factors of our function 𝑓 of 𝑥, we, therefore, have two factors. And we want to find the remaining two factors, which will help us find our remaining two zeros.

One method we can use here is to divide both sides of our equation by 𝑥 minus 𝑎 times 𝑥 minus 𝑏, the two factors that we have, as this will give us the other two factors multiplied together. Here, we’re gonna use a technique called polynomial division.

Instead of dividing our polynomial by one factor after another, which would be two steps, we can instead divide it by both at the same time. In order to do so, we’ll first need to multiply our two factors together.

To multiply these two factors together, we can first perform a rearrangement to make the calculation slightly easier. Working first on the left-hand factor, we distribute this negative sign throughout the brackets. We then do the same to our second factor.

When looking at these two factors, we may now become aware that we have a difference of two squares. A difference of two squares comes in the form 𝑎 plus 𝑏 times 𝑎 minus 𝑏. You may be aware that when multiplying out the difference of two squares. The two middle terms will cancel. And the solution will be 𝑎 squared minus 𝑏 squared.

Applying this to our factors, we see that we can set 𝑥 minus two as 𝑎 and the square root of three as 𝑏. We, therefore, see that multiplying our two factors together will give us 𝑥 minus two squared minus the square root of three squared. We now continue to simplify, first, by multiplying out 𝑥 minus two squared which is 𝑥 squared minus four 𝑥 plus four and then the square root of three squared which is just three. Finally, we are left with 𝑥 squared minus four 𝑥 plus one.

Now that we found this, we can divide our polynomial function by the newly found multiplication of our two factors. Polynomial division may look intimidating at first. But is actually very similar to long division which, will be a familiar tool.

We write out our polynomial with the divisor outside of the division symbol. We then compare both the leading term of our polynomial and our divisor by asking the question, how many times does 𝑥 squared fit inside two 𝑥 to the power of four? You should be able to see that 𝑥 squared multiplied by two 𝑥 squared is equal to two 𝑥 to the power of four. We can, therefore, write in this two 𝑥 squared above the division symbol in line with the correct power of 𝑥, which is the second power of 𝑥.

We now multiply the entire divisor by this two 𝑥 squared. And we find this is equal to two 𝑥 to the power of four minus eight 𝑥 cubed plus two 𝑥 squared. We now subtract this entire line from our polynomial, which is the dividend. Doing so, we find that we have negative 𝑥 cubed plus three 𝑥 squared plus three 𝑥 minus one remaining.

We now repeat the same process, comparing the leading term of our divisor and the remaining polynomial. We see that 𝑥 squared multiplied by negative 𝑥 is equal to negative 𝑥 cubed. We again write in this negative 𝑥 above the line in the place of the 𝑥 to the first power and multiply the negative 𝑥 by our entire divisor to find negative 𝑥 cubed plus four 𝑥 squared minus 𝑥.

We continue by subtracting this entire line from the remainder of our polynomial to get negative 𝑥 squared plus four 𝑥 minus one. We perform our process one more time and we see that 𝑥 squared multiplied by negative one is equal to negative 𝑥 squared.

We multiply our entire divisor by negative one to find negative 𝑥 squared plus four 𝑥 minus one. This is exactly the same as the polynomial that we have left over. And so by subtracting this entire line, we find that we have zero remaining. This is where our process terminates.

We have, therefore, found that dividing 𝑓 of 𝑥 by 𝑥 squared minus four 𝑥 plus one is equal to two 𝑥 squared minus 𝑥 minus one. In other words, 𝑓 of 𝑥 is the product of these two polynomials. And this can be seen by multiplying both sides of the equation by the denominator of the left-hand side.

Okay, let’s tidy things up to complete the final step of our problem. Looking at our expression, we may recall that our first set of brackets is derived from the two zeros given in the question. Let’s convert back to its original form to see the zeros more clearly.

In order to find the remaining two zeros associated with this set of brackets, we’ll need to factorize the polynomial inside the brackets. The coefficient of our 𝑥 squared term has only one positive factor pair, which is two and one. We can, therefore, write in two 𝑥 and 𝑥 as the first term of our brackets, respectively.

Our integral term negative one is the product of one and negative one. In order for our middle term to be negative 𝑥, the correct factorization is, therefore, two 𝑥 plus one multiplied by 𝑥 minus one. We can now write 𝑓 of 𝑥 as the product of the four following factors.

We now have the two zeros given by the question. And we know the other zeros exist, firstly, where two 𝑥 plus one is equal to zero, therefore where two 𝑥 is equal to negative one or where 𝑥 is equal to negative one over two, and finally where 𝑥 minus one is equal to zero or where 𝑥 is equal to one.

The four zeros of our polynomial are, therefore, two plus the square root of three, two minus the square root of three, negative one-half, and one.