Question

If √2 and -√2 are zeros of p(x)= 2x⁴+7x³-19x²-14x+30, find other zeros of p(x).​

Answer 1

Step-by-step explanation:

$p(x) = 2x {}^{4} + 7x {}^{3} - 19x {}^{2} - 14x + 30$

$g(x) = \sqrt{2} \: and \: - \sqrt{2}$

$g(x) = (x + \sqrt{2} )(x - \sqrt{2} )$

$g(x) = x {}^{2} - 2$

Now, we are dividing p(x) by g(x)

we get,

$p(x) = 2x {}^{4} + 7x {}^{3} - 19x {}^{2} - 14x + 30$

$g(x) = x {}^{2} - 2$

$q(x) = 2x {}^{2} + 7x - 15$

$r(x) = 0$

Now splitting the middle term of q(x),

we get,

$2x {}^{2} + 7x - 15$

$= 2x {}^{2} + 10x - 3x - 15$

$= 2x(x + 5) - 3(x + 5)$

$= (2x - 3)(x + 5) = 0$

So,

$x = \frac{3}{2}$

$x = - 5$

So, The all four zeroes of the polynomial are,

$= \sqrt{2} \\ = - \sqrt{2} \\ = \frac{3}{2} \\ = - 5$

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