F firefox0903 Oct 2012 3 0 Chicago Oct 28, 2012 #1 If 2x^{2}-dx+(31-d^{2})x+5 has a factor of x - d, what is the value of d if d is an integer? (Crying)

If 2x^{2}-dx+(31-d^{2})x+5 has a factor of x - d, what is the value of d if d is an integer? (Crying)

MarkFL Dec 2011 2,314 916 St. Augustine, FL. Oct 28, 2012 #2 Let: \(\displaystyle f(x)=2x^2-dx+(31-d^2)x+5\) If \(\displaystyle f(x)\) has a factor of \(\displaystyle x-d\) then \(\displaystyle f(d)=0\). Hint: use the rational roots theorem on the resulting cubic in \(\displaystyle d\).

Let: \(\displaystyle f(x)=2x^2-dx+(31-d^2)x+5\) If \(\displaystyle f(x)\) has a factor of \(\displaystyle x-d\) then \(\displaystyle f(d)=0\). Hint: use the rational roots theorem on the resulting cubic in \(\displaystyle d\).

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Oct 28, 2012 #3 post problems in the precalculus forum please.

F firefox0903 Oct 2012 3 0 Chicago Oct 28, 2012 #4 so if f(d)=0 then then the new f(x) is 2x^2+31x+5??

MarkFL Dec 2011 2,314 916 St. Augustine, FL. Oct 28, 2012 #5 No, that doesn't mean to let d = 0, that means when x = d, f(x) = 0.