Question

find the sum and the product of the zeros of PX is equal to 2 bracket x square - 3 bracket close plus x​

Answer 1

Answer:

px =2(x^2-3)+x

=2x^2-6+x

=2x^2+x-6

sum of the roots=-b/a

=-1/2

product of the roots=c/a

=-6/2

=-3

Answer 2

The sum of zeroes is -1/2 and product of zeros is -3

Solution:

Given that,

$p(x) = 2(x^2 - 3) + x\\\\Simplify\\\\p(x) = 2x^2 - 6 + x\\\\p(x) = 2x^2 + x - 6$

We have to find the zeros

Then,

$2x^2+x-6 = 0\\\\\text{Split the middle term }\\\\2x^2 - 3x +4x-6=0\\\\\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(2x^2-3x\right)+\left(4x-6\right) = 0\\\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\\\\mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\\\x\left(2x-3\right)+2\left(2x-3\right) = 0\\\\\mathrm{Factor\:out\:common\:term\:}2x-3\\\\\left(2x-3\right)\left(x+2\right) = 0$

$Therefore\\\\2x - 3 = 0\\\\x = \frac{3}{2}\\\\Also\\\\x+2=0\\\\x = -2$

$Thus\\\\\text{Sum of zeroes } = \frac{3}{2} - 2 = \frac{-1}{2} \\\\\ \text{Product of zeroes } = \frac{3}{2} \times -2 = -3$

Thus sum of zeroes is -1/2 and product of zeros is -3

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