The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, and f(2)=12, then what are x-intercepts of the graph f?

 

The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, and f(2)=12, then what are x-intercepts of the graph f?

 

\(f(x)=ax^3+bx^2+cx+d\)

 

\(f(0)=0\qquad so \qquad d=0\\ f(x)=ax^3+bx^2+cx\)

\(also\\ f(-1)=15,\;\;  f(1)=-5, \;\;and \;\; f(2)=12\)

 

so

 

\(f(-1)=-a+b-c=15\qquad(1)\\ f(1)=a+b+c = -5\qquad\quad(2)\\ f(2)=8a+4b+2c=12\qquad(3)\\ (1)+(2) 2b=10\\ b=5\\\)

Now we have

 

\(-a+5-c=15\qquad(1)\\ a+c=-10 \qquad \qquad (1b)\\ \\~\\ 8a+20+2c=12\qquad(3)\\ 4a+10+c=6 \\ 4a+c=-4 \qquad \qquad(3b)\\ \\~\\ (3b)-(1b)\\ 3a=6\\ a=2\\ c=-12 \)

 

 

\(f(x)=ax^3+bx^2+cx\\ f(x)=2x^3+5x^2-12x\\ f(x)=x(2x^2+5x-12)\\ f(x)=x(2x^2+8x-3x-12)\\ f(x)=x[2x(x+4)-3(x+4)]\\ f(x)=x[(2x-3)(x+4)]\\ \text{Roots are x = 0, 1.5 and -4}\)