Question

Find $-a+b-c+d$ if

$(x-\frac{1}{5})(x+1)(x+\frac{3}{7})$ $=a{x}^3+b{x}^2+cx+d$ for all real values of $x$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $4$

Answer 1

The correct option is A $0$

After solving the given equation in question we get

$\Rightarrow (x-\frac{1}{5})(x+1)(x+\frac{3}{7})$ = $35x^3+43x^2+5x-3$ and

$\Rightarrow 35x^3+43x^2+5x-3$ $=a{x}^3+b{x}^2+cx+d$

Now we have to compare the respective coefficients of $x,x^2,x^3$

By comparison, we get

$a=35$, $b=43$, $c=5$, $d=-3$

So the value of

$\Rightarrow -a+b-c+d$ $=-35+43-5-3=0$

Hence, option A is correct.