Question

Use the x-intercept method to find all real solutions of the equation. 14x^​3-53x^​2+41x-4=-4x^​3+x^​2+x+4

Answer 1

14x^3-4x^3-53x^2-x^2+41x-x-4-4=0  10x^3-54x^2+40x-8=0  2(5x-2)(x^2+5+2)=0  Divide both sides by 2   (5x-2)(x^2+5+2)=0  5x-2=0 5x=2 X=2/5  X^2+5+2=0 5+- sqrt (-5)^2-4×(1×2)/2×1  X= 5+- sqrt 17/2  So answer is   X= 5+- sqrt 17/2, 2/5

Answer 2

Answer:

The given polynomial function is

    $\rightarrow14x^3-53x^2+41x-4=-4x^3+x^2+x+4\\\\ \rightarrow14x^3-53x^2+41x-4+4x^3-x^2-x-4=0\\\\ \rightarrow18x^3-54x^2+40x-8=0\\\\\rightarrow 9x^3-27x^2+20x-4=0$

Now, we will plot graph of function on two dimensional coordinate plane

 The graph cuts x axis at three distinct points.So, there are three real solution of the polynomial expression.

  $x_{1}=0.333\\\\x_{2}=0.667\\\\x_{3}=2$