Answer: x = {1 , 5 , 7}
Step-by-step explanation:
We have x^3 - 13x^2 + 47x - 35 = 0.
Taking -13x^2, we can split it up into -x^2 - 12x^2
Taking 47x, we can split it up into 12x + 35x
Putting these into the original equation, we have:
x^3 - x^2 - 12x^2 + 12x + 35x - 35 = 0
From this you can see that we have 3 pairs of terms that we can easily factor. From the first pair we can factor x^2, from the second we can factor -12x, and from the third one we can factor 35.
1. (x^3 - x^2) = x^2 (x - 1)
2. (- 12x^2 + 12x) = -12x (x - 1)
3. (35x - 35) = 35 (x - 1)
Putting them back into the equation we have:
x^2 (x - 1) - 12x (x - 1) + 35 (x - 1) = 0
(x^2 - 12x + 35) (x - 1) = 0
To find the roots, we can set (x - 1) = 0 and (x^2 - 12x + 35) = 0.
x - 1 = 0
x = 1 (First root)
x^2 - 12x + 35 = 0
Factoring using the Sum-Product Rule:
Sum of -12, Product of 35 --> -7 and -5
x^2 - 12x + 35 = 0
(x - 7) (x - 5) = 0
We can see that x = 7 and x = 5.
So the solution set it x = {1 , 5 , 7}
The solution set will be {1,5,7}
Zero of a function is the root of the function for which the function value of that root will be zero.
If a is the zero of function f(x)=0 then f(a)=0.
So according to question,
the equation is
x³-13x²+47x-35=0
spilting -13x² as -x²-12x² and 47x as 12x+35x
⇒x³-x²-12x²+12x+35x-35=0
⇒x²(x-1)-12x(x-1)+35(x-1)=0
taking (x-1) common from each term
⇒(x-1)(x²-12x+35)=0
In another way, as 1 is zero of x³-13x²+47x-35=0
then (x-1) is one factor of x³-13x²+47x-35=0
another factor will be x³-13x²+47x-35/(x-1)=x²-12x+35
for which we can write the equation x³-13x²+47x-35=(x-1)(x²-12x+35).
⇒(x-1)(x²-5x-7x-35)=0
⇒(x-1)(x(x-5)-7(x-5))=0
⇒(x-1)(x-5)(x-7)=0
⇒x-1=0 or x-5=0 or x-7=0
⇒x=1 or x=5 or x=7
Therefore The solution set will be {1,5,7}
Learn more about the zero of the function
here: https://brainly.com/question/24380382
#SPJ2
