Question

Obtain all the zeroes of $x^3-7x+6,$ if one of its zero is 1.

Answer 1

Given:

To obtain all the zeroes of $x^3-7x+6,$ given that one of its zeroes is 1.

Calculations:

Let $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^3-7\mathrm{x}+6$

${\mathrm{x}}^3-7\mathrm{x}+6=(\mathrm{x}-1)({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c})$

Comparing the coefficient of $x^2$:

$$\begin{gathered} 0=a+b \\ \Rightarrow a=b\left(1\right) \end{gathered}$$

Comparing the coefficient of $x$:

$-7=c-b\left(2\right)$

Comparing the constant term:

$$\begin{gathered} 6=-c \\ \Rightarrow c=-6\left(3\right) \end{gathered}$$

Put (3) in (2)

$$\begin{gathered} \mathrm{b}=1=\mathrm{a}(\mathrm{from}(1\left)\right) \\ \therefore {\mathrm{x}}^2+\mathrm{x}-6=(\mathrm{x}+3)(\mathrm{x}-2) \end{gathered}$$

Final Answer:

Thus, the other zeroes of p (x) are -3, 2 .