The factored type is the product of constants and two-dimensional terms, that is the root of a function as well as the graph's x-intercepts, and the further calculation can be defined as follows:
Given:
[tex]\bold{f(x)=x^3+5x^2+4x-6}[/tex]
To find:
factored form=?
Solution:
[tex]\bold{f(x)=x^3+5x^2+4x-6}[/tex]
putting the value x=-1,-2,-3...... to find the function value 0.
[tex]\to \bold{f(-1)=-1^3+5(-1)^2+4(-1)-6}\\\\[/tex]
[tex]\bold{=-1+5\times 1+4 \times (-1)-6}\\\\ \bold{= -1+5-4-6} \\\\ \bold{=-6}\\\\[/tex]
[tex]\to \bold{f(-2)=-2^3+5(-2)^2+4(-2)-6}[/tex]
[tex]\bold{=-8+5\times 4+4 \times(- 2)-6}\\\\\bold{= -8+20-8-6}\\\\\bold{=-2}\\\\[/tex]
[tex]\to \bold{f(-3)=-3^3+5(-3)^2+4(-3)-6}[/tex]
[tex]\bold{=-27+5\times 9+4 \times(- 3)-6}\\\\\bold{= -27+45-12-6}\\\\\bold{=0}\\\\[/tex]
Therefore, [tex]\bold{(x+3)}[/tex] is the factor of the function.
Let [tex]\bold{(x+3)}[/tex] divide by the function:
[tex]\to \bold{\frac{x^3+5x^2+4x-6}{(x+3)}= x^2+2x-2}[/tex]
compare the value [tex]\bold{x^2+2x-2}[/tex] by the standard equation [tex]\bold{ax^2+bx+c=0}[/tex] so,
[tex]\bold{a=1}\\\\\bold{b=2}\\\\\bold{c=-2}\\\\[/tex]
Putting value into the quadratic equation formula:
[tex]\bold{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}[/tex]
[tex]\bold{x=\frac{-(2) \pm \sqrt{(2)^2-4\times 1 \times -2}}{2\times 1}}\\\\\bold{x=\frac{-2 \pm \sqrt{4+8}}{2}}\\\\\bold{x=\frac{-2 \pm \sqrt{12}}{2}}\\\\\bold{x=\frac{-2 \pm 2\sqrt{3}}{2}}\\\\\bold{x=\frac{2(-1 \pm \sqrt{3})}{2}}\\\\\bold{x=(-1 \pm \sqrt{3})}\\\\\bold{x=(-1 + \sqrt{3}) (-1 -\sqrt{3})}\\\\[/tex]
Adding the factor value that is [tex]\bold{(x+3)}[/tex] so, the value is "[tex]\bold{(x+3)(-1+\sqrt{3})(-1-\sqrt{3}) }[/tex]"
So, the final answer is "[tex]\bold{f(x)=(x+3)(x-(-1+\sqrt{3}))(x-(-1-\sqrt{3}))}[/tex]"
Learn more:
brainly.com/question/16754415
