Question

find all other zeros of polynomial x^4+x^3-23x^2-3x+60, if it is given that two of its zeros are √3 and -√3



pls also show steps​

Answer 1

Step-by-step explanation:

Given :-

√3 and -√3 are the zeroes of x⁴+x³-23x²-3x+60

To find :-

Find all other zeros of polynomial ?

Solution :-

Given bi-quadratic polynomial=x⁴+x³-23x²-3x+60

Let P(x) = x⁴+x³-23x²-3x+60

Given zeroes = √3 and -√3

We know that by Factor theorem

If √3 is a zero of p(x) then (x-√3) is a factor of p(x)

If -√3 is a zero of p(x) then (x+√3) is a factor of p(x)

If √3 and -√3 are the zeroes of p(x) then

=>(x-√3)(x+√3) is a factor of p(x).

=> x²-(√3)²

Since (a+b)(a-b) = a²-b²

=> x²-3 is the factor of p(x)

To get remaining zeroes we have to divide p(x) by (x-√3)(x+√3)

x⁴+x³-23x²-3x+60 ÷ (x-√3)(x+√3)

=> x⁴+x³-23x²-3x+60 ÷ (x²-3)

x²-3) x⁴+x³-23x²-3x+60 (x² + x - 20

x⁴ -3x²

(-) (+)

_______________

0 + x³ -20x²-3x

x³ -3x

(-) (+)

________________

0 -20x² +0 +60

-20x² +60

(+) (-)

__________________

0

__________________

The quotient = x² +x -20

=> x²+5x-4x-20

=> x(x+5)-4(x+5)

=> (x+5)(x-4)

P(x) = x⁴+x³-23x²-3x+60

=> P(x) = (x²-3)(x+5)(x-4)

=> P(x) = (x+√3)(x-√3)(x+5)(x-4)

To get zeroes we write p(x) = 0

=> (x+√3)(x-√3)(x+5)(x-4) = 0

=> x+√3 = 0 or x-√3 = 0 or x+5 = 0 or x-4 = 0

=> x = -√3 or √3 or -5 or 4

Answer:-

All the Zeroes of the given bi-quadratic polynomial is √3,-√3 ,-5 and 4

Used formulae:-

• (a+b)(a-b) = a²-b²

• Long Division

• To get zeores of a polynomial we equate it to zero.