Question

Let $f\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d,$ where $a,b,c,d\in Q.$ If $f\left(\sqrt{3}\right)=f(3+i)=0,$ then the value of $a+b+c+d$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $-9$
• B. $11$
• C. $-11$
• D. $9$

Answer 1

The correct option is C $-11$

$f\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$
As $a,b,c,d\in Q,$ complex roots and irrational roots occur in conjugate pairs, so the roots of the equation $f\left(x\right)=0$ are
$\sqrt{3},-\sqrt{3},3+i,3-i$
Therefore,
$f\left(x\right)=(x-\sqrt{3})(x+\sqrt{3})[x-(3+i\left)\right][x-(3-i\left)\right]$

Now, we know that
$f\left(1\right)=1+a+b+c+d\Rightarrow (1-\sqrt{3})(1+\sqrt{3})(-2-i)(-2+i)=1+a+b+c+d\Rightarrow 1+a+b+c+d=-2\left(5\right)\therefore a+b+c+d=-11$