Question

$\text{Let}f\left(x\right)=x^6–2x^5+x^3+x^2–x–1$ and $g\left(x\right)=x^4–x^3–x^2–1$ be two polynomials. $\text{Let}a,b,c\text{and}d\text{the roots of}g\left(x\right)=0$. Then the value of $f\left(a\right)+f\left(b\right)+f\left(c\right)+f\left(d\right)$ is

• A. $-5$
• B. $0$
• C. $4$
• D. $5$

Answer 1

The correct option is B $0$

$f\left(x\right)=x^6–2x^5+x^3+x^2–x–1$ and $g\left(x\right)=x^4–x^3–x^2–1$
We know $a,b,c$ and $d$ are roots of
$x^4–x^3–x^2–1=0$
$\text{Now using this and substituting it in}f\left(x\right)$
$f\left(x\right)=x^6–2x^5+x^3+x^2–x–1\Rightarrow f\left(x\right)=x^2(x^4–x^3–x^2–1)-x^5+x^4+x^3+2x^2-x-1\Rightarrow f\left(x\right)=x^2(x^4–x^3–x^2–1)-x(x^4–x^3–x^2–1)+2x^2-2x-1\Rightarrow f\left(x\right)=(x^2-x)(x^4–x^3–x^2–1)+2x^2-2x-1$
$f\left(a\right)+f\left(b\right)+f\left(c\right)+f\left(d\right)=2(a^2+b^2+c^2+d^2)-2(a+b+c+d)-(1+1+1+1)=2\left\{\right(\sum a{)}^2-2\sum ab\}-2\sum a-4=2\{1^2-2(-1)\}-2-4=0$