find all the zeroes of the polynomial. 2x4 + 5x3 - 11x2 + 20x + 12, if two of its zeroes are 2 and -2. I dont want links that have answers similar to questions like this. I want answer for this question only without any link.
Answers
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.
Answer:
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