+ b + c is equal to zero and b + c is equal to zero then find the zeros of the given polynomial x 4 + b x cube + 3 x square + bx + c
Answers
If 1 and 3 are two zeros of polynomial x3-ax2-13x+b=0, then what is the value of a and b?
I really like straight-forward logic!
If x = 1 and x = 3 are two of the solutions of f(x) = 0
then f(x) must equal (x – 1)(x – 3)(x – c) where c is the 3rd solution.
—————————————————————————————-
f(1) must be zero! So 1 – a – 13 + b = 0
This simplified is: – a + b = 12 ----------------------------EquA
—————————————————————————————
f(3) must also be zero! So 27 – 9a – 39 + b = 0
This simplified is: – 9a + b = 12 ---------------------------EquB
————————————————————————————-
EquA – EquB: 8a = 0 so a = 0
Subs in EquA: b = 12
______________________________________________________
The above is all you asked for but to continue for FUN……..
596 Views ·
Why is SBI replacing the traditional debit cards having the magnetic strips with the new cards ha...
If 1 and 3 are two zeros of polynomial x3-ax2–13x+b=0, then what is the value of a and b?
As a cubic equation, the polynomial has three roots (although they do not have to be distinct). At present, we only know two of these roots, 1 and 3. Let us call the third root P.
We can express the equation as (x−1)(x−3)(x−P)=0
Expanding this, we get (x2−4x+3)(x−P)=0
Expanding further, we get:x3−(P+4)x2+(4P+3)x−3P=0
But we already know that the equation is x3−ax2−13x+b=0
Equating the coefficients of these two equations, we have:
x1: −13=4P+3⇒4P=−16⇒P=−4
x2: −a=−(P+4)⇒a=P+4⇒a=0
x0: b=−3P⇒b=12
So the required answer is: a=0 and b=12
