Question

Determine all factors of the expression 3x^3+7x^2-18x+8 if one of the factors is x-1

Answer 1

Answer:

The factors are x-2,$x-\frac{2}{3}$ and x+4

Therefore the expression can be written as $3x^3+7x^2-18x+8=(x-2)(x-\frac{2}{3})(x+4)$

Step-by-step explanation:

Given expression is $3x^3+7x^2-18x+8$

And also given that x-1 is one of the factors

i.e.,x-1=0

 x=1

To find the factors equate the given expression to zero

$3x^3+7x^2-18x+8=0$

Using synthetic division to find the factors

1_|    3       7       -18        8

       0       3        10       -8

   ___________________

       3      10       -8         0

Therefore the quadratic equation is  $3x^2+10x-8=0$

To find the factors of the above equation:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a and b are coefficients of $x^2$ and x respectively

$x=\frac{-10\pm\sqrt{10^2-4(3)(-8)}}{2(3)}$ Where a=3 , b=10 and c=-8

$=\frac{-10\pm\sqrt{100+96}}{6}$

$=\frac{-10\pm\sqrt{196}}{6}$

$=\frac{-10\pm14}{6}$

$x=\frac{-10\pm14}{6}$

Therefore $x=\frac{-10+14}{6}$  and $x=\frac{-10-14}{6}$

$x=\frac{4}{6}$  and $x=\frac{-24}{6}$

Therefore $x=\frac{2}{3}$  and $x=-4$

Therefore the factors are x-2,$x-\frac{2}{3}$  and x+4

$3x^3+7x^2-18x+8=(x-2)(x-\frac{2}{3})(x+4)$