Write the function as the product of (x+3), (x+4), and another expression of degree 2 with no real roots.
Example Solution: y=x^4+7x^3+13x^2+7x+12
Practice makes perfect
We want to find a quartic function with the given x-values as its only real zeros.
x=- 3 and x=- 4To do so, we can write the function as the product of (x+3), (x+4), and another expression of degree 2 with no real roots — for example, ( x^2+1 ).
y=(x+3)(x+4)( x^2+1 )
Let's simplify the above by using the Distributive Property.
We found that y=x^4+7x^3+13x^2+7x+12 is a quartic function with - 3 and - 4 as its only real zeros.
Alternative Solution
Solving the exercise by using transformations of a graph.
If you like graphs, there is another way of solving this exercise. Consider the parent function y=x^4.
Since the difference between the given zeros is one, we have to find a quartic function where the roots are 1 unit apart, and having them centered at 0 is the easiest. Then we can translate the graph over to the correct place using transformations. Let's find a quartic with zeros at ± 12, because that will add up to 1. c|c
Distance between & Distance between
-3 and -4 & - 12 and 12 [0.8em]
-3-(-4)=1 & 12- ( - 12 ) =1
To do so, we need to translate the parent function ( 12)^4= 116 units down. Therefore, we need to consider the function y=x^4- 116.
Finally, - 4 and - 3 are 72 units to the left of - 12 and 12, respectively.
This means that we need to translate the last function 72 units to the left. Now, we need to consider the function y=( x+ 72 )^4- 116.
We can see above that the quartic function with - 3 and - 4 as its only real zeros is y=( x+ 72 )^4- 116. Note that, although this function is different than the one we found using the algebraic method, it is also a correct answer.
