Question

Find the zeroes of the polynomial f(x)=x3-12x2+47x-60,if it is given that sum of its two zeroes is
9

Answer 1

Answer:

4,5,10

Step-by-step explanation:

x³-12x²+47x-60

α+β+γ=-(-12)/1=12

9+γ=12

y=12-9=3

αβγ=-(-60)/1=60

3αβ=60

αβ=60/3=20

α+β=9

α+αβ/α=9

α+20/α=9

α²-9α+20=0

α²-5α-4α+20=0

α(α-5)-4(α-5)=0

(α-5)(α-4)=0

α=5,THEN β=20/5=4

SO THE ZEROS ARE

4,5,10

Answer 2

Zeroes of the polynomial f(x) = x³ -12x² + 47x - 60 are 3 , 4 and 5 if sum of its two zeroes is 9

Given:

f(x) = x³ -12x² + 47x - 60

sum of its two zeroes is 9

To Find:

Zeroes

Solution:

f(x) = x³ -12x² + 47x - 60

Let say zeroes are α , β , γ

Sum of  two zeroes = α + β = 9

Sum of all zeroes = α + β + γ = - (-12)/1  = 12

Substitute α + β = 9

9 + γ = 12

=> γ = 3

Hence one zero is 3

Product of zeroes = αβγ = -(-60)/1 = 60

Substitute γ = 3

=> αβ3 = 60

=> αβ = 20

α + β = 9

αβ = 20

Quadratic Equation will be x² - 9x + 20 = 0

Using middle term split

(x - 4)(x - 5) = 0

Hence α and β will be  4 and 5

So zeroes are 3 , 4 , 5

Verification:

f(x) = x³ -12x² + 47x - 60

f(3) = 3³ -12(3)² + 47(3) - 60 = 0

f(4) = 4³ -12(4)² + 47(4) - 60 = 0

f(5) = 5³ -12(5)² + 47(5) - 60 = 0

Hence Zeroes of the polynomial f(x) = x³ -12x² + 47x - 60 are 3 , 4 and 5 if sum of its two zeroes is 9