Question

the sum of all the zeros of the polynomial p(x)=5x⁵-20x⁴+5x³+50x²-20x-8 is __________ and product is ___________.​

Answer 1

Answer:

sun of zeroes = -coefficient of X/ coefficient of x^2

=-20/50

Answer 2

Answer:

Answer

5.0/5

5

Deleted account

Answer:

The zeroes are 2, -2, 3 and -1/2.

Step-by-step explanation:

Put f(x) = 2x⁴ + 5x³ - 11x² + 20x + 12.

Then checking that 2 and -2 are zeroes:

f(2) = 80 and f(-2) = 80.

So there is a mistake in the question. Both of these are out by 40x, so it appears the question should have said "-20x". Let's work from that then.

Put f(x) = 2x⁴ + 5x³ - 11x² + 20x + 12.

Checking that 2 and -2 are zeroes:

f(2) = 0 and f(-2) = 0. Good!

This means that x-2 and x+2 are factors. Therefore (x-2)(x+2) = x² - 4 is a factor. Hence

f(x) = 2x⁴ + 5x³ - 11x² + 20x + 12 = ( x² - 4 ) ( ax² + bx + c )

for some a, b, c.

Equating coefficients of x⁴: 2 = a.

Equating constant coefficients: 12 = -4c => c = -3.

Equating coefficients of x: 20 = -4b => b = -5.

So ax² + bx + c = 2x² - 5x - 3 = ( 2x + 1 ) ( x - 3 ) we have:

2x⁴ + 5x³ - 11x² + 20x + 12 = ( x - 2 ) ( x + 2 ) ( 2x + 1 ) ( x - 3 )

The other zeroes are therefore -1/2 and 3.