# Question Video: Writing Polynomials as a Product of Linear and Irreducible Quadratic Factors and Listing All Zeros Mathematics

Consider 𝑓(𝑥) = 𝑥⁴ + 3𝑥³ − 5𝑥² − 3𝑥 + 4. Write 𝑓(𝑥) as the product of linear and irreducible quadratic factors. List all zeros of 𝑓(𝑥).

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### Video Transcript

Consider 𝑓 of 𝑥 is equal to 𝑥 to the power of four plus three 𝑥 cubed minus five 𝑥 squared minus three 𝑥 plus four. Write 𝑓 of 𝑥 as the product of linear and irreducible quadratic factors. List all zeros of 𝑓 of 𝑥.

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 𝑓 of 𝑥 by substituting in values of 𝑥. If our answer is equal to zero, then the number we’ve substituted in is a root.

We will start by considering 𝑓 of one, substituting 𝑥 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 — one, three, and four — gives us eight. And adding the two negative terms — negative five and negative three — gives us negative eight. As eight minus eight is equal to zero, we can say that 𝑓 of one is equal to zero. As one is a root, 𝑥 minus one is a factor of 𝑓 of 𝑥.

If this was a cubic equation, we would then divide our cubic function by 𝑥 minus one. However, as this is a quartic, we will repeat this process to find a second root of 𝑓 of 𝑥. We’ll now consider 𝑓 of negative one, substituting in 𝑥 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 𝑓 of minus one is equal to zero, we can say that 𝑥 plus one is a factor. As 𝑥 minus one and 𝑥 plus one are both factors of 𝑓 of 𝑥, their product will also be a factor of 𝑓 of 𝑥. We can expand or multiply out these two parentheses using the FOIL method.

Multiplying the first terms gives us 𝑥 squared. Multiplying the outside terms gives us positive 𝑥. The inside terms multiply to give negative 𝑥. And finally, the last terms multiply to give us negative one. 𝑥 minus 𝑥 is equal to zero. Therefore, 𝑥 minus one multiplied by 𝑥 plus one is equal to 𝑥 squared minus one.

We can now divide our quartic function 𝑓 of 𝑥 by 𝑥 squared minus one. We can do this following the methods of division, multiplication, and then subtraction. Firstly, we divide 𝑥 to the power of four by 𝑥 squared. This is equal to 𝑥 squared. We then multiply 𝑥 squared by 𝑥 squared minus one. This is equal to 𝑥 to the power of four minus 𝑥 squared. We then subtract our two lines. 𝑥 to the four minus 𝑥 to the four is equal to zero. Three 𝑥 cubed minus zero is three 𝑥 cubed. And negative five 𝑥 squared minus negative 𝑥 squared is negative four 𝑥 squared.

We drop down the next term in our initial expression and repeat the process. Three 𝑥 cubed divided by 𝑥 squared is equal to three 𝑥. Multiplying three 𝑥 by 𝑥 squared minus one gives us three 𝑥 cubed minus three 𝑥. We then subtract the two lines once again. This gives us negative four 𝑥 squared, as three 𝑥 cubed minus three 𝑥 cubed is zero and negative three 𝑥 minus negative three 𝑥 is also zero.

We drop down the positive four term and repeat the process a third time. Negative four 𝑥 squared divided by 𝑥 squared is equal to negative four. Multiplying negative four by 𝑥 squared minus one gives us negative four 𝑥 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 𝑓 of 𝑥, 𝑥 to the power of four plus three 𝑥 cubed minus five 𝑥 squared minus three 𝑥 plus four, is equal to 𝑥 squared minus one multiplied by 𝑥 squared plus three 𝑥 minus four. We already know that 𝑥 squared minus one factorizes to 𝑥 minus one multiplied by 𝑥 plus one. Factorizing the second quadratic, 𝑥 squared plus three 𝑥 minus four, gives us 𝑥 plus four multiplied by 𝑥 minus one. We’ve now written 𝑓 of 𝑥 as the product of four linear factors.

We notice at this stage though that two of the factors are identical. They’re 𝑥 minus one. This means that we can rewrite 𝑓 of 𝑥 as 𝑥 plus one multiplied by 𝑥 plus four multiplied by 𝑥 minus one all squared. 𝑓 of 𝑥 written as the product of linear and irreducible quadratic factors is 𝑥 plus one multiplied by 𝑥 plus four multiplied by 𝑥 minus one all squared.

The second part of our question asked us to list all zeros of 𝑓 of 𝑥. This is another way of saying write all the roots of 𝑓 of 𝑥. In our initial working, we established that 𝑥 equals minus one and 𝑥 equals positive one were roots of 𝑓 of 𝑥. The third possible root comes from the factor 𝑥 plus four and gives us an answer of 𝑥 equals negative four. This means that there are three zeros of 𝑓 of 𝑥. They are negative one, negative four, and positive one.