Question

If $p\left(x\right)=x^2-4x+3,\mathrm{then}p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Let p(x) = x² – 4x + 3

$$\begin{gathered} p\left(2\right)={\left(2\right)}^2-4\left(2\right)+3 \\ =4-8+3 \\ =-1 \\ p\left(-1\right)={\left(-1\right)}^2-4\left(-1\right)+3 \\ =1+4+3 \\ =8 \\ p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+3 \\ =\frac{1}{4}-2+3 \\ =\frac{1}{4}+1 \\ =\frac{1+4}{4} \\ =\frac{5}{4} \\ \mathrm{Now}, \\ p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=-1-8+\frac{5}{4} \\ =-9+\frac{5}{4} \\ =\frac{-36+5}{4} \\ =-\frac{31}{4} \end{gathered}$$

Hence, if $p\left(x\right)=x^2-4x+3,\mathrm{then}p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\underline{-\frac{31}{4}}.$