Factorize $x^{3} + 3 x^{2} - 28 x$ and verify the relationship between the zeroes and coefficients.

Open in App

Solution

$x^{3} + 3 x^{2} - 28 x$

= x(x+7)(x-4)

So the zeroes are -7, 4 and 0

$α + β + γ$ = -7 + 4 + 0 = -3 = $\frac{- b}{a}$

$α β + β γ + γ α$ = -28 + 0 + 0 = -28 = $\frac{c}{a}$

$α β γ$ = 0 = $\frac{- d}{a}$

Suggest Corrections

Similar questions

Factorize $3 x^{2} - (3 \sqrt{3} - 1) x - \sqrt{3}$ and verify relationship between the zeroes and the coefficients. [2 MARKS]

Factorize $6 t^{2} + 13 t + 7$ and verify the relationships between the zeroes and the coefficients.

Factorize $z^{2} + 3 \sqrt{3} z - 12$ and verify the relationship between zeroes and coefficient [2 MARKS]

3 x^{2} - 2

1, - 1,

- 3

x^{3} + x^{2} - x - 3

Join BYJU'S Learning Program