Verify that the numbers given along side of the cubic polynomials below are their zero.
(i) $f (x) = 2 x^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2$
(ii) $g (x) = x^{3} - 4 x^{2} + 5 x - 2; 2, 1, 1$

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Solution

(i) $f (x) = {2 x}^{3} + x^{2} - 5 x + 2$
$f (1 / 2) = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2, f (1) = 2 + 1 - 5 + 2 = 0, f (- 2) = - 16 + 4 + 10 + 2 = 0$
So, $1 / 2$ , $+ 1$ , $- 2$ are zeroes of $f (x)$
(ii) $g (x) = x^{3} - {4 x}^{2} + 5 x - 2$
$g (2) = 8 - 8 \times 2 + 10 - 2 = 0, g (1) = 1 - 4 + 5 - 2 = 0$

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Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in in each case:

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2 x^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2

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Verify that the numbers given along side of the cubic polynomials below are their zero. (i) f(x)=2x3+x2−5x+2;12,1,−2(ii) g(x)=x3−4x2+5x−2;2,1,1

(i) f(x)=2x3+x2−5x+2f(1/2)=14+14−52+2,f(1)=2+1−5+2=0,f(−2)=−16+4+10+2=0So, 1/2, +1, −2 are zeroes of f(x)(ii) g(x)=x3−4x2+5x−2g(2)=8−8×2+10−2=0,g(1)=1−4+5−2=0

Verify that the numbers given along side of the cubic polynomials below are their zero.
(i) $f (x) = 2 x^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2$
(ii) $g (x) = x^{3} - 4 x^{2} + 5 x - 2; 2, 1, 1$