Chapter 2, Problem 100CR

a.

To determine

To find: all zeros of the function.

Answer to Problem 100CR

  $-5,-5,i,i.$

Explanation of Solution

Given information:

  $f(x)=x^4+10x^3+26x^2+10x+25$

Concept used:-

Rational root test:- Let f(x) be a polynomial and c be constant term in f(x) and a is leading coefficient of f(x) then ratio of possible divisors of constant term and leading coefficient can rational zeros of function.

Rational root = $\frac{divisors\text{ of constant term }}{divisors\text{ of leading cofficient}}$

Calculation:- Possible divisors of constant term are $\pm 1,\pm 5,\pm 25$ and divisor of leading coefficient term is 1

1^stmethod:-

Since, we know zeros of a polynomial satisfies polynomial so, by putting these numbers in polynomial we can check which number is zero of polynomial.

2^ndmethod: - using synthetic division method we can also check the roots of polynomial.

Now, we check, is (x+5) factor of $x^4+10x^3+26x^2+10x+25$ .

  

Hence, (x+ 5) is factor of $x^4+10x^3+26x^2+10x+25$ or -5 is zero of $x^4+10x^3+26x^2+10x+25$ .

So,

  $x^4+10x^3+26x^2+10x+25$

  $=(x+5)(x^3+5x^2+x+5)$

  $\begin{array}{l}(x^3+5x^2+x+5)\\ =x^2(x+5)+1(x+5)\\ =(x^2+1)(x+5)\end{array}$

  $x^4+10x^3+26x^2+10x+25=(x+5)(x^3+5x^2+x+5)$ Or

  $x^4+10x^3+26x^2+10x+25=(x+5)(x^2+1)(x+5)$ Or

  $x^4+10x^3+26x^2+10x+25={(x+5)}^2(x^2+1)$ ------ (1)

Applying shreedhracharya for $(x^2+1)$ to get zeros

  $\begin{array}{l}x=\frac{0\pm \sqrt{0-4}}{2}\\ x=\frac{\pm 2i}{2}=\pm i\end{array}$

So, $(x^2+1)=(x+i)(x-i)$ .

Therefore, (1) becomes

  $x^4+10x^3+26x^2+10x+25={(x+5)}^2(x+i)(x-i)$ ------- (2)

So, zeros of function are $-5,-5,i,i.$

b.

To determine

To represent: the polynomial as product of linear factor form.

Answer to Problem 100CR

  $(x+5)(x+5)(x+i)(x-i)$

Explanation of Solution

Given information:

Zeros of given function are $-5,-5,i,i.$

Calculation:

From part (A) and equation (2), we have

  $x^4+10x^3+26x^2+10x+25=(x+5)(x+5)(x+i)(x-i)$

c.

To determine

the x- intercept of function and verify it using graphing ultity.

Answer to Problem 100CR

  $x=-5$

Explanation of Solution

Given information:

From part (B) we have

  $x^4+10x^3+26x^2+10x+25=(x+5)(x+5)(x+i)(x-i)$

Concept used:-

The graph of polynomial cuts the x- axis at its real zeros.

Calculation:-

Since, we have

  $x^4+10x^3+26x^2+10x+25=(x+5)(x+5)(x+i)(x-i)$

Real zeros of polynomial is -5.

So graph of given function will cut x-axis at x= -5

Verification:-

  

Using graph we can see graph of function cuts x-axis at x = -5.