need help with these problems pls CHAPTER 4 NAME TEST FORM B

need help with these problems pls

CHAPTER 4 NAME TEST FORM B For each polynomial function: a Find the rational and then the other zeros; that is, solve f(x) 0 b) Factor f(x) into linear factors. 15. f(x) x 2 x 5x-10 16. f(r) x 2x3 46x2 98x -147 is.

Solution

15. f(x) = x^3 +2x^2 -5x -10

The factor of the leading coefficient (1) is 1 .The factors of the constant term (-10) are 1 2 5 10 . Then the Rational Roots Tests yields the following possible solutions:

±1/1, ±2/1, ±5/1, ±10/1

Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(2)=0.

So, use factor theorem to find remaining roots:

(x^3 +2x^2 -5x -10)/(x +2) = x^2-5

x^2 -5 =0 ---> x = sqrt5 , -sqrt5

So, zeros of f(x) are x= -2 , sqrt5 , -sqrt5

f(x) = (x +2)(x -sqrt5)( x +sqrt5)

16. f(x) = x^4 +2x^3+46x^2 +98x -147

This polynomial has no rational roots that can be found using Rational Root Test.

We have to use quartic formulas :

a X⁴   + bX ³   + cX² + dX + e = 0

find : f = c - (3b²/8) = 44.5

find : g = d + (b³ / 8) - (b*c/2) = 53

find : h = e - (3*b⁴/256) + (b ² * c/16) - ( b*d/4) =  -184.6875

Now

we \"plug\" the numbers \'f\', \'g\' and \'h\' into the following cubic equation:

Y³ + (f/2)*Y² + ((f² -4*h)/16)*Y -g²/64 = 0

solve the cubic equation , we get : y1 = 0.249 ;   y2 = -11.25 + i* 7.00 ;

y3 =   -11.25 - i* 7.0

Let \'p\' and \'q\' be the square roots of ANY 2 non-zero roots (Y₁ Y₂ or Y₃).

p= 1 +i*3.5

q=  1 -i*3.5

r= -g/(8*pq) = -0.5

s= b/(4*a) = 0.5

X₁= p + q + r -s = 1
X₂= p - q - r -s = i7
X₃= -p + q - r -s = -i7
X₄= -p - q + r -s = -3

f(x) = (x -1)(x+3)(x^49)