Question

find the zeroes of x^3+11x^2+23x-35 if one of its zeroes is-7​

Answer 1

Step-by-step explanation:

Method 1 :-

Factorise -

x³ + 11x² + 23x - 35

= x³ + 7x² + 4x² + 28x - 5x - 35

= x²(x + 7) + 4x(x + 7) - 5(x + 7)

= (x + 7)(x² + 4x - 5)

= (x + 7)(x² - x + 5x - 5)

= (x + 7)[x(x - 1) + 5(x - 1)]

= (x + 7)(x + 5)(x - 1)

Zeroes are -

x + 7 = 0 , x + 5 = 0 , x - 1 = 0

• x = - 7, - 5, 1

Method 2 :-

Divide -

If - 7 is zero of polynomial x³ + 11x² + 23x - 35, then (x + 7) is the factor.

x + 7)x³ + 11x² + 23x - 35(x² + 4x - 5

x³ + 7x²

(-) (-)

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4x² + 23x - 35

4x² + 28x

(-) (-)

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-5x - 35

-5x - 35

(+) (+)

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X X

Now,

Factorising x² + 4x - 5

x² + 4x - 5

= x² - x + 5x - 5

= x(x - 1) + 5(x - 1)

= (x + 5)(x - 1)

Zeroes are -

x + 5 = 0 and x - 1 = 0

• x = - 5 and x = 1

Answer 2

$\bold{\red{\underline{\underline{\rm{ Given :}}}}}$

${x}^{3} + 11 {x}^{2} + 23x - 35$

$\bold{\blue{\underline{\underline{\rm{ To\:Find :}}}}}$

Zeros of this equation.

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$\green{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

Factorize the equation →

$= {x}^{3} + 7 {x}^{2} + 4 {x}^{2} + 28x - 5x - 35 \\ = {x}^{2} (x + 7) + 4x(x + 7) - 5(x + 7) \\ = (x + 7)( {x}^{2} + 4x - 5) \\ = (x + 7)( {x}^{2} + 5x - x - 5) \\ = (x + 7)(x(x + 5) - 1(x + 5)) \\ = (x + 7)(x - 1)(x + 5)$

So the zeros are -

$x + 7 = 0 \\ x = - 7$

$x - 1 = 0 \\ x = 1$

$x + 5 = 0 \\ x = - 5$

So the zeros are -7 , -5 , 1 .

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