Question

A cubic polynomial $f( x )= ax ^3+ bx ^2+ cx + d$ has a graph which is tangent to the $x$-axis at 2 , has another $x$-intercept at -1 , and has y-intercept at -2 as shown. The value of, $a + b + c + d$ equals

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (B)

The polynomial must be of the form $f(x)=a(x-2)^2(x$ $+1)$ because it has a double zero at 2 and a zero at -1 .
To solve for ' $a$ ', note that $f (0)= a (0-2)^2(0+1)=-2$.
It follows that $a =-1 / 2$ Hence $f(x)=-\frac{1}{2}\left[x^3-3 x^2+4\right]$
$\therefore a =-\frac{1}{2}, b =\frac{3}{2}, c =0 ; d =-2$
$ \Rightarrow c + b + c + d =-1$