Question

find the zeroes of 2x³-5x²-14x+8

Answer 1

here is your answer.. Mark it as brilliant

Answer 2

$The\ zeros\ of\ 2x^3-5x^2-14x+8 \quad :\quad x=-2,\:x=\frac{1}{2},\:x=4$

Solution:

From given,

$2x^3-5x^2-14x+8\\\\substitute\ x = -2\\\\2(-2)^3 - 5(-2)^2 - 14(-2) + 8 \\\\-16 - 20 + 28 + 8\\\\-36 + 36\\\\0$

Thus,

x = -2 is a zero

Now the expression becomes:

$(x+2)(2x^2-9x+4)$

Factorize the second $2x^2-9x+4$

$2x^2-9x+4\\\\Break\:the\:expression\:into\:groups\\\\\left(2x^2-x\right)+\left(-8x+4\right)\\\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-x\mathrm{:\quad }\\\\\mathrm{Factor\:out\:}-4\mathrm{\:from\:}-8x+4\mathrm{:\quad }$

$x\left(2x-1\right)-4\left(2x-1\right)\\\\\mathrm{Factor\:out\:common\:term\:}2x-1\\\\\left(2x-1\right)\left(x-4\right)$

$2x - 1 = 0\\\\x = \frac{1}{2}\\\\x - 4 = 0\\\\x = 4$

Thus the zeros are:

$x=-2,\:x=\frac{1}{2},\:x=4$

Learn more:

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