nth degree polynomial function with rear coefficients satisfying the given conditions
- Thread starter jlg
- Start date
\(\displaystyle f(x)=a(x-r_{1})(x-r_{2})\cdot\cdot\cdot(x-r_{n})\) where "a" is your leading coefficient and the "r's" are your roots.
So for your problem you have: \(\displaystyle f(x)=a(x+3)(x-\frac{1}{4})(x-2i)(x+2i)\)
Mutliplying all this out you get (check my calcs to be sure
): \(\displaystyle f(x)=a(x^4+\frac{11}{4}x^3+\frac{13}{4}x^2+11x-3)\)
Now they also give you \(\displaystyle f(2)=-144\). This allows you to solve for "a". Plug in x=2 into the above equation, set it equal to -144 and solve for "a". Again, check my calculations, but ou get \(\displaystyle a=\frac{-36}{23}\).
So your final answer is \(\displaystyle f(x)=\frac{-36}{23}(x^4+\frac{11}{4}x^3+\frac{13}{4}x^2+11x-3)\)
So for your problem you have: \(\displaystyle f(x)=a(x+3)(x-\frac{1}{4})(x-2i)(x+2i)\)
Mutliplying all this out you get (check my calcs to be sure
): \(\displaystyle f(x)=a(x^4+\frac{11}{4}x^3+\frac{13}{4}x^2+11x-3)\)Now they also give you \(\displaystyle f(2)=-144\). This allows you to solve for "a". Plug in x=2 into the above equation, set it equal to -144 and solve for "a". Again, check my calculations, but ou get \(\displaystyle a=\frac{-36}{23}\).
So your final answer is \(\displaystyle f(x)=\frac{-36}{23}(x^4+\frac{11}{4}x^3+\frac{13}{4}x^2+11x-3)\)