Find a line with x-intercept 34 and another with x-intercept - 34.
Example Solution: t(x)=16x^3-9x and s(x)=16x^4-9x^2. Explanation: See solution.
Practice makes perfect
Let's build third and fourth degree polynomials, starting with linear functions.
Linear Functions
There is no linear function with two x-intercepts, but we can find one with x-intercept 34.
f(x)=4x-3
We can check that x= 34 is indeed a zero.
f(3/4)=4(3/4)-3=3-3=0
Similarly, we can find a function with a zero at x=- 34.
g(x)=4x+3
Quadratic Functions
Notice that if we multiply the two linear functions we found, we get a quadratic function with both 34 and - 34 as zeros.
h(x)=(4x-3)(4x+3)
Substituting x= 34 makes the first part of the expression 0,
substituting x=- 34 makes the second part of the expression 0.
Either way, the product is 0.
Let's expand the product to find the standard form of the expression.
We found a quadratic function with zeros at ± 34.
Third Degree Polynomial
Because of the Zero Product Property, if we multiply the quadratic we found with any function, we can only add to the set of zeros.
To get a third degree polynomial, we can choose any linear multiplier. Let's choose x.
t(x)=x* h(x)=x(16x^2-9)=16x^3-9x
This gives a third degree polynomial function that has zeros at ± 34.
Fourth Degree Polynomial
Similarly, we can get a fourth degree polynomial function that has zeros at ± 34 if we multiply h(x)=16x^2-9 by any second degree polynomial.
Let's choose x^2.
s(x)=x^2* h(x)=x^2(16x^2-9)=16x^4-9x^2
Extra
Solutions with zeros only at ± 34.
In our explanation we found polynomial functions with zeros at ± 34 and also at 0.
To get functions where the only zeros are at ± 34, we need to be careful how we choose the added factors so as not to introduce additional zeros.
Let's look at all possibilities.
Third Degree Polynomials
Factor Form
Expanded Form
(4x-3)^2(4x+3)
64 x^3 - 48 x^2 - 36 x + 27
(4x-3)(4x+3)^2
64 x^3 + 48 x^2 - 36 x - 27
Fourth Degree Polynomials
Factor Form
Expanded Form
(4x-3)^3(4x+3)
256 x^4 - 384 x^3 + 216 x - 81
(4x-3)^2(4x+3)^2
256 x^4 - 288 x^2 + 81
(4x-3)(4x+3)^3
256 x^4 + 384 x^3 - 216 x - 81
The possible third and fourth degree polynomial functions with zeros only at ± 34 are constant multiples of the ones in the tables above.
There are also fourth degree polynomial functions with real zeros at ± 34 and two complex zeros.
f(x)=(4x-3)(4x+3)(ax^2+bx+c), where b^2-4ac< 0
The condition b^2-4ac<0 tells that the discriminant of the quadratic factor is negative, which guarantees that the zeros are not real.
