Question

the ans which satisfies would me marked as brainliest..

find the cubic polynomial whose zeroes are -3, -1, and 4 ..
find the zeroes of 5x² + 2x

solve these..
plz...frnds it's urgent !!

Answer 1

Hi ...dear..

here is your solutions..
..
.hope this helped you...
Regards,#shubhendu

Answer 2

Given:

Three zeroes which are 3, 1/2 and -1

To find:

The cubic polynomial whose zeroes are 3, 1/2, and -1.

So,

Let

\alpha=3α=3\beta= \frac{1}{2}β= 21\gamma = (-1)γ=(−1)

So,

We know that,

If α, β, γ are the zeroes of the cubic polynomial, then the polynomial will be

k(\boxed{\red{\sf x^{3}-(sum\ of\ zeroes)x^{2}+(product\ of\ zeroes\ two\ taken\ at\ time)x - (product\ of\ zeroes)}})k( x 3

−(sum of zeroes)x 2

+(product of zeroes two taken at time)x−(product of zeroes) )

Which is

k(\boxed{\red{\sf x^{3}-(\alpha+\beta+\gamma)x^{2}+(\alpha \beta + \beta \gamma + \gamma \alpha)x - (\alpha \beta \gamma )}})k(

x 3−(α+β+γ)x 2 +(αβ+βγ+γα)x−(αβγ) )

Where 'k' is a constant.

So,

Sum of zeroes

= α + β + γ

= 3+\frac{1}{2} + (-1)3+ 21 +(−1)

= \frac{6+1-2}{2} 26+1−2

= \frac{7-2}{2} 27−2

= \frac{5}{2} 25

Now,

αβ +βγ + γα

= [3*\frac{1}{2}]+[\frac{1}{2}*(-1)] + [(-1)*3][3∗ 21]+[ 21∗(−1)]+[(−1)∗3]

= [\frac{3}{2}]+[\frac{-1}{2}] + [-3][ 23 ]+[ 2−1 ]+[−3]

= \frac{3+(-1)-6}{2} 23+(−1)−6

= \frac{3-7}{2} 23−7

= \frac{-4}{2} 2−4

= -2

Also,

αβγ

= 3 * \frac{1}{2}* (-1)3∗ 21 ∗(−1)

= \frac{-3}{2} 2−3

Now,

The cubic polynomial is

\boxed{\red{\sf x^{3}-(\frac{5}{2})x^{2}+(-2)x - (\frac{-3}{2} )}} x 3−( 25 )x 2 +(−2)x−( 2−3 )

= \red{\sf x^{3}-\frac{5}{2}x^{2}-2x +\frac{3}{2} }x 3−25x 2 −2x+ 23

= \red{\sf \frac{2x^{3}-5x^{2}-4x +3}{2} } 2

2x 3−5x 2 −4x+3

= 2x³ - 5x² - 4x + 3

It is the required cubic polynomial.