Question

Obtain all the zeroes of the polynmial x⁴ + 6x³ + x² - 24x + 20 if two of its zeroes are 2 and -5

Grade 10, Polynomials.

Answer 1

Correction to question:-

$\boxed{\sf{P(x) =x^{4}+6x^{3}+x^{2}-24x-20}}$

Two of zeroes are 2 and (-5)

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According to FACTOR THEOREM

(x-2) and (x+5) are factors are p(x)

Also,

$(x - 2)(x + 5) \\ \\ = {x}^{2} + 5x - 2x - 10 \\ \\ = {x}^{2} + 3x - 10$

x²+3x-10 is also a factor of p(x)

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Now,

Diving p(x) by x²+3x-10

We get :-

Q(x) = x² + 3x +2

r(x) = 0

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We have :-

$p(x) = ({x}^{2} + 3x -10)( {x}^{2} + 3x + 2) \\ \\ \\ = (x - 2)(x + 5)( {x}^{2} + x + 2x + 2) \\ \\ \\ = (x - 2)(x + 5)(x + 1)(x + 2)$

$\rule{200}{2}$

Thus

Zeroes are as follows :-

$x - 2 = 0 \\ \\ x = 2$

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

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$x + 1 = 0 \\ \\ x = ( - 1)$

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$x + 2 = 0 \\ \\ x = ( - 2)$

$\rule{200}{2}$

Answer 2

Answer: -(√17 + 3)/2 and   (√17 - 3)/2

Step-by-step explanation:

Given,

P(x): x⁴ + 6x³ + x² -24x + 20

Let the other two zeroes be α and β

Comparing p(x) With the standard form of  equation of  4th degree

ax⁴ + bx³ + cx² + dx + E,

a = 1

b = 6

c = 1

d = -24

E = 20 ,

Now,

We know that,

Sum of  zeroes = -b/a

2  + (-5) + α + β =  -6/1

2 - 5 +  α + β = -6

-3 +  α + β = -6

α + β = -6 + 3

α + β = -3 .....i)

Product  of zeroes = E/a

2 × (-5) ×  α × β = 20/1

-10αβ = 20

αβ = -20/10

αβ = -2

We know  that,

(α - β)² = (α + β)² - 4αβ

(α - β)² = (-3)² -4(-2)

(α - β)² =  9 + 8

(α - β)² = 17

α - β =  √17 ...........ii)

Adding equation i) and ii)

2α = √17 - 3

α = (√17 - 3)/2

Subtracting equation ii) from i)

2β = -3 - √17

2β = -(√17 + 3)

β = -(√17 + 3)/2