Question

Write polynomial $P$ as a product of linear factors:$P\left(x\right)=9x^3-19x-10$

Answer 1

$p\left(x\right)=9x^3-19x-10$

To find the zeroes of this polynomial, let us try $-1,1$ as trial error.

$\Rightarrow p\left(1\right)=9{\left(1\right)}^3-19-10\ne 0$

$\Rightarrow p(-1)=9{(-1)}^3-19\left(1\right)-10=0$

Hence, $(x+1)$ will be a factor of $p\left(x\right).$

To find factors we can divide the $p\left(x\right)$ by $(x+1),$

Using long division method

So, the quotient $=9x^2-9x-10$

To find the other factors, use common factor theorem,

$9x^2-9x-10$

$=9x^2+6x-15x-10$

$=3x(3x+2)-5(3x+2)$

$=(3x-5)(3x+2).$

Hence, the polynomial $p\left(x\right)$ can be written as $p\left(x\right)=(x+1)(3x-5)(3x+2).$

Hence, the answer is $(x+1)(3x-5)(3x+2).$