Question:
Solution:
Consider the following polynomial function:
[tex]h(x)=2x^3+23x^2+58x-35[/tex]Since all coefficients are integers, we can apply the rational zeros theorem.
the rational zeros theorem states:
If h(x) is a polynomial with integer coefficients and if P/Q is a zero of h(x), then P is a factor of the constant term of h(x) and Q is a factor of the leading coefficient of h(x).
Now, the trailing coefficient (the coefficient of the constant term) is -35.
its factors (with the plus sign and the minus sign) are:
[tex]\pm1,\text{ }\pm5,\text{ }\pm7,\text{ }\pm35[/tex]These are the possible values for P.
The leading coefficient (the coefficient of the term with the highest degree) is 2.
its factors (with the plus sign and the minus sign) are:
[tex]\pm1,\text{ }\pm2[/tex]These are the possible values for Q.
Thus, all possible values of P/Q are:
[tex]\frac{P}{Q}\colon\text{ }\pm\frac{1}{1}\text{ , }\pm\frac{1}{2}\text{ , }\pm\frac{5}{1},\text{ }\pm\frac{5}{2}\text{ , }\pm\frac{7}{1},\text{ }\pm\frac{7}{2},\text{ }\pm\frac{35}{1},\text{ }\pm\frac{35}{2}[/tex]now simplify and remove the duplicates (if any). These are the possible rational roots:
[tex]\pm1,\text{ }\pm\frac{1}{2}\text{ , }\pm5,\text{ }\pm\frac{5}{2},\text{ }\pm7,\text{ }\pm\frac{7}{2},\text{ }\pm35,\text{ }\pm\frac{35}{2}[/tex]Next, check the possible roots: if a is a root of the polynomial h(x), the remainder from the division of h(x) by x-a should equal 0. According to the remainder theorem, this means that
[tex]h(a)=0[/tex]Check 1: divide h(x) by x-1:
[tex]h(1)=48[/tex]thus, the remainder is 48.
Check -1: divide h(x) by x+1:
[tex]h(-1)=-72[/tex]
thus, the remainder is -72.
Check 1/2: divide h(x) by x-(1/2):
[tex]h(\frac{1}{2})=0[/tex]
thus, the remainder is 0. Hence 1/2 is a root.
Check -1/2: divide h(x) by x+(1/2):
[tex]h(-\frac{1}{2})=-\frac{117}{2}[/tex]
thus, the remainder is -117/2.
Check 5: divide h(x) by x-5:
[tex]h(5)=1080[/tex]
thus, the remainder is 1080.
Check -5: divide h(x) by x+5:
[tex]h(-5)=0[/tex]
thus, the remainder is 0. Hence -5 is a root.
Check 5/2: divide h(x) by x-(5/2):
[tex]h(\frac{5}{2})=285[/tex]
thus, the remainder is 285.
Check -5/2: divide h(x) by x+(5/2):
[tex]h(-\frac{5}{2})=-\frac{135}{2}[/tex]
thus, the remainder is -135/2.
Check 7: divide h(x) by x-7:
[tex]h(7)=2184[/tex]
thus, the remainder is 2184.
Check -7: divide h(x) by x+7:
[tex]h(-7)=0[/tex]
thus, the remainder is 0. Hence -7 is a root.
Check 7/2: divide h(x) by x-(7/2):
[tex]h(\frac{7}{2})=\frac{1071}{2}[/tex]
thus, the remainder is 1071/2.
Check -7/2: divide h(x) by x+(7/2):
[tex]h(-\frac{7}{2})=-42[/tex]
thus, the remainder is -42.
Check 35: divide h(x) by x-35:
[tex]h(35)=115920[/tex]
thus, the remainder is 115920.
Check -35: divide h(x) by x+35:
[tex]h(-35)=-59640[/tex]
thus, the remainder is -59640.
Check 35/2: divide h(x) by x-(35/2):
[tex]h(\frac{35}{2})=\frac{37485}{2}[/tex]
thus, the remainder is 37485/2.
Finally, check -35/2: divide h(x) by x+(35/2):
[tex]h(-\frac{35}{2})=-4725[/tex]
thus, the remainder is -4725.
We can conclude that the correct answer is:
The zeros of the given function are:
[tex]\frac{1}{2},\text{ }-5,\text{ -7}[/tex]
