Question

2)
(0) Let p(x)=
$x { }^{2} - 4x + 3.$
Find the value of po).p(1),p(2), p(3) and obtain zeroes of the
polynomial p(x).
(i) Check whether -3 and 3 are the zeroes of the polynomial x? -9.

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Answer 1

${\huge{\underline{\sf {\purple{Question : }}}}}$

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Let p(x) = $\sf {x}^{2} - 4x + 3$. Find the values of p(0), p(1), p(2), p(3) and obatain the zeroes of the polynomial p(x).

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Check whether -3 and 3 are the zeroes of the polynomial $\sf {x}^{2} - 9$

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${\huge{\underline{\sf {\purple{Answer : }}}}}$

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Part 1 :

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To obtain the values of p(0), p(1), p(2), p(3) we need to put the values 0, 1, 2, 3 in place of x in p(x).

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$\sf \: p(x) = {x}^{2} - 4x + 3$

$\:$

$\sf \: p(0) = {0}^{2} - 4(0) + 3$

⇢ $\sf \: p(0) = 3$

$\:$

$\sf \: p(1) = {1}^{2} - 4(1) + 3$

⇢ $\sf \: p(1) = 1 - 4 + 3$

⇢ $\sf \: p(1) = 4 - 4$

⇢ $\sf \: p(1) = 0$

$\:$

$\sf \: p(2) = {2}^{2} - 4(2) + 3$

⇢ $\sf \: p(2) = 4 - 8 + 3$

⇢ $\sf \: p(2) = 7 - 8$

⇢ $\sf \: p(2) = - 1$

$\: \:$

$\sf \: p(3) = {3}^{2} - 4(3) + 3$

⇢ $\sf \: p(1) = 9 - 12 + 3$

⇢ $\sf \: p(3) = 12 - 12$

⇢ $\sf \: p(3) = 0$

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Since p(1) and p(3) = 0, so, they are the zeroes of the polynomial p(x).

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The given polynomial can have only 2 zeroes since the degree (highest power) of the polynomial is 2.

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Part 2 :

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To check if -3 and 3 are the zeroes of the polynomial we need to put the values in place of x in the given polynomial and if the answer is 0 then the given number is the zero of the polynomial.

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Let,

$\sf \: f(x) = {x}^{2} - 9$

$\sf \: f( - 3) = { (- 3)}^{2} - 9$

↬ $\sf \: f( - 3) = 9 - 9$

↬ $\sf \: f( - 3) = 0$

$\: \:$

$\sf \: f( 3) = {3}^{2} - 9$

↬ $\sf \: f( 3) = 9 - 9$

↬ $\sf \: f( 3) = 0$

Since, f(-3) and f(3) = 0 so, they are the zeroes of the polynomial.