Question

two zeros of P(X) is equal to x^3 6x^2+ 11 x - 6 are 2 and 3. find the third zero​

Answer 1

Step-by-step explanation:

Answer:

The third zero is 1

Step-by-step explanation:

Given 2 and 3 are the zeroes of polynomial x^3-6x^2+11x-6x3−6x2+11x−6

Thus, (x - 2)(x - 3) will be a factor of x^3-6x^2+11x-6x3−6x2+11x−6

Let the third factor be (ax+b)

∵ Given polynomial is cubic so,

x^3-6x^2+11x-6=(x -2)(x - 3)(ax + b)x3−6x2+11x−6=(x−2)(x−3)(ax+b)

x^3-6x^2+11x-6=(x^2-3x-2x+6)(ax + b)x3−6x2+11x−6=(x2−3x−2x+6)(ax+b)

x^3-6x^2+11x-6=(x^2-5x+6)(ax + b)x3−6x2+11x−6=(x2−5x+6)(ax+b)

x^3-6x^2+11x-6=(ax^3+bx^2-5ax^2-5bx+6ax+6b)x3−6x2+11x−6=(ax3+bx2−5ax2−5bx+6ax+6b)

x^3-6x^2+11x-6=(ax^3+(b-5a)x^2+(6a-5b)x+6b)x3−6x2+11x−6=(ax3+(b−5a)x2+(6a−5b)x+6b)

Comparing coefficients on both sides

a = 1 and b = - 1

Thus, the factor (ax + b) = (x - 1)

Hence, the third zero is x = 1

Answer 2

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Correct Question:-

• Two zeros of P(X) is equal to x^3 6x^2+ 11 x - 6 are 2 and 3. find the third zero?

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Answer:- 1 is third zero of

this equation

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Step by step explanation:-

$⇢ {x}^{3} - {6x}^{2} + 11x - 6 \\ \\ ⇢ {x}^{3} - {x}^{2} - {5}^{2} + 5x + 6x - 6 \\ \\ ⇢ {x}^{2} (x -1) - 5x(x - 1) + 6(x - 1) \\ \\ ⇢(x - 1) ( {x}^{2} - 5x + 6) \\ \\ ⇢(x - 1)( {x}^{2} - 2x - 3x + 6) \\ \\ ⇢(x - 1)(x(x - 2) - 3(x - 2)) \\ \\ ⇢(x - 1)(x - 2)(x - 3) - - - - - (i)$

Now solve (i) equation for zeros of this equation.

⇢ x-1=0

⇢x=1

⇢x-2=0

⇢x=2

⇢x-3=0

⇢x=3

So , zeros of this equation are 1, 2 and 3.

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How I solve this....

• First factorise the equation.
• Solve the last factors of this equation for finding zeros of any equation.

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