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(i) \(2x^3 + x^2 - 5x + 2 ; {{1}\over{2}},1,-2\)

(ii) \(x^3 - 4x^2 + 5x - 2 ; 2,1,1\)

Answer :

(i)The general form of cubic equation is \(ax^3 + bx^2 + cx + d\).

Here, \(a = 2,b = 1,c = -5,d = 2\)

On substituting the value of zeroes in the given equation, p(x)= \(2x^3 + x^2 - 5x + 2\)

p\(({{1}\over{2}})\)= \(2({{1}\over{2}})^3 + ({{1}\over{2}})^2 + -5({{1}\over{2}}) + 2\)

=>\({{1}\over{4}} + {{1}\over{4}} - {{5}\over{2}} + 2 = 0 \)

p\((1)\)= \(2(1)^3 + (1)^2 + -5(1) + 2\)

=>\(2 + 1 - 5 + 2 = 0 \)

p\((-2)\)= \(2(-2)^3 + (-2)^2 + -5(-2) + 2\)

=>\(-16 + 4 + 10 + 2 = 0 \)

Hence, these zeroes satisfies the the given equation.

Now, we'll check whether the zeroes satisfies the following equations:

Let the three zeroes be p,q and r

\(p + q + r\) = \({{-b}\over{a}}\)............(i)

\(pq + qr + rp\) = \({{c}\over{a}}\)..........(ii)

\(pqr\) = \({{-d}\over{a}}\).................(iii)

On checking equation (i):=> \({{1}\over{2}} + 1 + (-2)\)

=>\({{3}\over{2}} - 2 \)

=> \({{-1}\over{2}} = {{-b}\over{a}}\)

On checking equation (ii):=> \({{1}\over{2}} × 1 + 1 × (-2) + (-2) × {{1}\over{2}}\)

=>\({{1}\over{2}} - 2 - 1\)

=> \({{-5}\over{2}} = {{c}\over{a}}\)

On checking equation (iii):=> \({{1}\over{2}} × 1 × (-2)\)

=>\({{-2}\over{2}}\)

=> \( -1 = {{-d}\over{a}}\)

**Hence it is verified the numbers given alongside of the cubic polynomials are their zeroes.**

**(ii)** The general form of cubic equation is \(ax^3 + bx^2 + cx + d\).

Here, \(a = 1,b = -4,c = 5,d = -2\)

On substituting the value of zeroes in the given equation, p(x)= \(x^3 - 4x^2 + 5x - 2 \)

p\((2)\)= \(2^3 - 4(2)^2 + 5(2) - 2 \)

=>\(8 - 16 + 10 - 2 = 0 \)

p\((1)\)= \(1^3 - 4(1)^2 + 5(1) - 2\)

=>\(1 - 4 + 5 - 2 = 0\)

**Hence, these zeroes satisfies the the given equation.**

Now, we'll check whether the zeroes satisfies the following equations:

Let the three zeroes be p,q and r

\(p + q + r\) = \({{-b}\over{a}}\)............(i)

\(pq + qr + rp\) = \({{c}\over{a}}\)..........(ii)

\(pqr\) = \({{-d}\over{a}}\).................(iii)

On checking equation (i):=> \(2 + 1 + 1\)

=>\(4\)

=> \({{-(-4)}\over{1}} = {{-b}\over{a}}\)

On checking equation (ii):=> \(2 × 1 + 1 × 1 + 1 × 2\)

=>\(2 + 1 + 2 = 5\)

=> \({{5}\over{1}} = {{c}\over{a}}\)

On checking equation (iii):=> \(2 × 1 × 1\)

=>\(2\)

=> \( {{-(-2)}\over{1}}= {{-d}\over{a}}\)

**Hence it is verified the numbers given alongside of the cubic polynomials are their zeroes.**

- Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time and the product of its zeroes are \(2, -7,-14\) respectively.
- If the zeroes of the polynomial \(x^3 - 3x^2 + x + 1\) are \(a-b , a , a+b\) find \(a\) and \(b\).
- If the two zeroes of the polynomial \(x^4- 6x^3 - 26x^2 -138x - 35\) are \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\) find other zeroes.
- If the polynomial \(x^4- 6x^3 + 16x^2 -25x + 10\) is divided by another polynomial \(x^2 -2x + k \) the remainder comes out to be \(x +a\) find \(k\) and \(a\).

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