Question

Show that: $\mathrm{x}+3$ is a factor of $69+11𝑥-{\mathrm{x}}^2+{\mathrm{x}}^3.$

Answer 1

Use Factor theorem

To prove: $\mathrm{x}+3$ is a factor of $69+11𝑥-{\mathrm{x}}^2+{\mathrm{x}}^3.$

According to factor theorem, if $\mathrm{f}\left(\mathrm{x}\right)$ is a polynomial of degree $\mathrm{n}\ge 1$ and $\text{'}\mathrm{a}\text{'}$ is any real number then $(\mathrm{x}-\mathrm{a})$ is a factor of $\mathrm{f}\left(\mathrm{x}\right)$ provided $\mathrm{f}\left(\mathrm{a}\right)=0$.

Proof:

Assign a function to the given polynomial, we get

Let us assume $\mathrm{f}\left(\mathrm{x}\right)$$=$$69+11𝑥-{\mathrm{x}}^2+{\mathrm{x}}^3$

Put $\mathrm{x}=-3$ in $\mathrm{f}\left(\mathrm{x}\right)$$=$$69+11𝑥-{\mathrm{x}}^2+{\mathrm{x}}^3$, we get

$\mathrm{f}(-3)$$=$$69+11\times (-3)-{(-3)}^2+{(-3)}^3$

$=$$69-33-9-27$

$=$ $36-36$

$=0$

∵ $\mathrm{f}(-3)$$=0$

Thus, by factor theorem $\mathrm{x}+3$ is the factor $69+11𝑥-{\mathrm{x}}^2+{\mathrm{x}}^3$.

Hence proved.