1.

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

Answer»

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

(-) (-)

____________________

4x² + 23x - 35

4x² + 28x

(-) (-)

____________________

-5x - 35

-5x - 35

(+) (+)

____________________

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


Discussion

No Comment Found

Related InterviewSolutions