From the given four options, another factor of polynomial [tex]function of x=2 x^{3}+x^{2}-25 x+12[/tex] is x = -4.
Solution:
Given that x = 3 is a zero of the polynomial function [tex]x=2 x^{3}+x^{2}-25 x+12[/tex]
Need to check if any other zeros is present in four given option. Let us understand what is zero of a polynomial. A zero for polynomial is that value of variable x for which dependent function of (x) = 0.
So we need to check for all four values of x,
Corresponding value of polynomial [tex]function of x=2 x^{3}+x^{2}-25 x+12[/tex] and if it is equal to 0 that that value of x is zero of f of x.
On substituting x= -3, we get
[tex]\begin{array}{l}{2(-3)^{3}+(-3)^{2}-25(-3)+12} \\\\ {\Rightarrow -54+9+75+12=42}\end{array}[/tex]
As function of -3 = 42, which is not equal to 0,
So x= -3 is not a zero of polynomial [tex]\rightarrow 2 x^{3}+x^{2}-25 x+12[/tex]
On substituting x = 4, we get
[tex]\begin{array}{l}{\rightarrow 2(4)^{3}+(4)^{2}-25(4)+12} \\\\ {\Rightarrow (2 \times 64)+16-100+12} \\\\ {\Rightarrow 128+16+12-100=56}\end{array}[/tex]
As function of 4 = 56, which is not equal to 0,
So x= 4 is not a zero of polynomial [tex]\rightarrow 2 x^{3}+x^{2}-25 x+12[/tex]
On substituting x = -4, we get
[tex]\begin{array}{l}{\rightarrow 2(-4)^{3}+(-4)^{2}-25(-4)+12} \\\\ {\Rightarrow (2 \times(-64))+16+100+12} \\\\ {\Rightarrow -128+16+12+100=-128+128=0}\end{array}[/tex]
As function of -4 = 0 , so x = -4 is a zero of a zero of polynomial [tex]function of x=2 x^{3}+x^{2}-25 x+12[/tex]
On substituting x = 12, we get
[tex]\begin{array}{l}{function of 12=2(12)^{3}+(12)^{2}-25(12)+12} \\\\ {\Rightarrow function of 12=1728+144-300+12} \\\\ {\Rightarrow function of 12=1584}\end{array}[/tex]
As function of 12 = 1584, which is not equal to 0,
So, x= 12 is not a zero of polynomial [tex]function of x=2 x^{3}+x^{2}-25 x+12[/tex]
Therefore, the another zero is x=-4.
