Finding zeros as a product of linear factors?

Question

Ok...so for this I am given: f(x)= x^3+11x^2+39x+29 and they want me to find the zeros and write the polynomial as a product of linear factors. We just started this chapter today and no one has any idea how to do this!

Answer 1

The roots of a function occur for values of x such that f(x)=0.

Startng with

f(x) = x³ + 11x² + 39x + 29

we want the values of x where

x³ + 11x² + 39x + 29 = 0

This is a polynomial of degree 3 (the highest power in the polynomial) which means there are up to 3 values of x that will satisfy this equation. You can rewrite it as

(ax+b) (cx+d) (ex+f) = 0

where each factor on the left is a linear function in x. The equation can only be satisfied when one of these factors itself is 0 (recall that a*b = 0 only if a=0 and/or b=0).

As the others mentioned, one of those values was found by inspection, f(-1) = 0. If f(-1)=0 then one root is x=-1 or x+1=0. This is a linear function in x.

Now factor the (x+1) from the equation and you get

(x + 1) (x² + 10x + 29) = 0

Using the quadratic equation, you find that

x = {-10 ± √[10² - 4(1)(29)]}/(2*1)
x = [-10 ± √(100 - 116)]/2
x = ½[-10 ± √-16]
x = ½[-10 ± 4i]
x = -5 ± 4i

Recall that i=√-1

Thus one root is x = -5 + 4i and another is x = -5 - 4i

x = -5 + 4i gives the linear function x + 5 - 4i = 0
x = -5 - 4i gives the linear function x + 5 + 4i = 0

So now we know that

f(x) = (x + 1)(x + 5 - 4i)(x + 5 + 4i)

and the roots are given by

(x + 1)(x + 5 - 4i)(x + 5 + 4i) = 0

Answer 2

Start by taking a guess. By inspection, I noticed that x = -1 is a zero. That is to say, f(-1) = 0. Therefore, (x + 1) is a linear factor of the polynomial. If r is a root of the polynomial, (x - r) is a factor of it. Now, by further inspection, it looks like I can write the original polynomial as (x + 1)(x^2 + 10x + 29). The discriminant of the quadratic polynomial factor is 10^2 - 4*29 = 100 - 116 = -16, which is negative, meaning that the roots are complex and not useful in a factorization. Thus, the original polynomial cannot be written as the product of three linear factors; it cannot be factored any further than I already have, and x = -1 is the only real root.

Answer 3

to find the zeroes, plug in the factors of leading coefficient and the factors of the constant for x

leading coefficient is 1. the factors are +/- 1
the constant is 29. The factors are +/-29

try -1

(-1)^3 + 11(-1)^2 + 39(-1) + 29
-1 + 11 - 39 + 29
10 - 10
0

-1 is one of the roots. so (x + 1) is one of the factors

to find the other roots. Use sythetic long division. Since x + 1 is in the form x - k, thus, k is -1

-1 l 1 ... 11 ... 39 ... 29
....l ....... -1 ... -10 ... -29
--------------------------------
.... 1 .... 10 ... 29 ... 0

so you have
(x + 1) (x^2 + 10x + 29)

since you can not factor the rest. (x^2 + 10x+ 29) has no roots. The only zero is -1

Answer 4

considering 29 is a significant extensive style, there are basically 4 achieveable aspects: (x + a million), (x - a million), (x + 29), (x - 29) do away with (x - a million) and (x - 29), considering all coefficients in f(x) are useful. making use of long branch: x + a million is going into x^3 + 11x^2 + 39x + 29, x^2 circumstances x^3 + 11x^2 + 39x + 29 - (x^3 + x^2) 10x^2 + 39x + 29 LEFT OVER x + a million is going into 10x^2 + 39x + 29, 10x circumstances 10x^2 + 39x + 29 - (10x^2 + 10) 29x + 29 LEFT OVER x + a million is going into 29x + 29, 29 circumstances 29x + 29 - (29x + 29) 0 LEFT OVER x + a million is a element x^3 + 11x^2 + 39x + 29 = (x + a million)(x^2 + 10x + 29) Now, we could element x^2 + 10x + 29: a = a million b = 10 c = 29 x = (-10 +/- sqrt(10^2 - 4(a million)(29)))/2(a million) x = (-10 +/- sqrt(a hundred - 116))/2 x = (-10 +/- sqrt(-sixteen))/2 x = (-10 +/- 4i)/2 x = -5 +/- 2i NO actual answer basically one x-intercept: x = -a million x^3 + 11x^2 + 39x + 29 = (x + a million)(-5 + 2i)(-5 - 2i) 0's: x = -2, 5 - 2i, 5 + 2i