Question

Find the zeroes of the polynomial p(x) =x3+4x +x-6 if iy gives that the product of two of its zeroes is 6

Answer 1

X³+4x²+x-6

a = 1, b = 4, c = 1, d = -6

→ Sum of zeroes = -b/a = -4

→ Sum of the product of zeroes taken two at a time = c/a = 1

→ Product of zeroes = -d/a = -(-6)/1 = 6

Answer 2

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Here is your answer.
${x}^{3} + 4 {x}^{2} + x - 6 = 0 \\ \\ x {}^{3} + 2 {x}^{2} + 2 {x}^{2} + x - 6 = 0 \\ \\ {x}^{2} (x + 2) + (2 {x}^{2} + 4x - 3x - 6) = 0 \\ \\ {x}^{2} (x + 2) + (2x(x + 2) - 3(x + 2)) = 0 \\ \\ {x}^{2} (x + 2) + (x + 2)(2x - 3) = 0 \\ \\ (x + 2)( {x}^{2} + 2x - 3) = 0 \\ \\( x + 2)( {x}^{2} + 3x - x - 3) = 0 \\ \\ (x + 2)(x(x + 3) - 1(x + 3)) = 0 \\ \\ (x + 2)(x + 3)(x - 1) = 0 \\ \\ x = - 2 \: \\ x = - 3 \: and \\ x = 1$
The product of Two zeros are (-2)×(-3) = 6

Third zeros is 1

Therefore zeros are 1 , - 2, and -3

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