Question

Hii guys

Obtain all other zeroes of the polynomial x^4+6x³+x²-24x-20 if two other zeroes are 2 and -5

Answer 1

Note :- if a & b are zeros of Any Polynomial f(x) , than (x - a)(x - b) is completely divide the polynomial f(x).

Solution :-

Given that, 2 & (-5) are the two zeros of polynomial.

So,

→ (x-2)(x+5)

→ x² + 5x - 2x - 10

→ x² + 3x - 10 is completely divide the given Polynomial .

So,

x² + 3x - 10 )x⁴ + 6x³ + x² -24x - 20( x²+3x + 2

x⁴+ 3x³- 10x²

(-ve) 3x³ + 11x² - 24 x

3x³ + 9x² - 30x

(-ve) 2x² + 6x - 20

2x² + 6x - 20

(-ve) 0.

So,

→ x² + 3x + 2 = 0

→ x² + 2x + x + 2 = 0

→ x(x + 2)+1(x+2) = 0

→ (x + 2)(x + 1) = 0

Putting Both Equal to Zero we get,

→ x = (-2) & (-1). (Ans).

Hence, All 4 Zeros of Given Polynomial are [(-5),(-2),(-1) & 2 ]..

Answer 2

Answer:

• Polynomial : x⁴ + 6x³ + x² - 24x - 20
• Zeroes : 2 and - 5

If 2 and - 5 are zeroes of the Polynomial then, Polynomial will be Divisible by (x - 2) and (x + 5), and so to there Products.

↠ (x - 2)(x + 5)

↠ x² + 5x - 2x - 10

↠ x² + 3x - 10

• (x² + 3x - 10) will completely Divide Polynomial. So by Division.

$\rule{150}{1}$

$\boxed{\begin{array}\quad\begin{tabular}{m{3.5em}ccccc}&&\bf x^2& \bf+3x&\bf+2\\\cline{1-7}\multicolumn{2}{l}{\sf x^2+\text{3x - 10}\big)}&x^4&+6x^3&+x^2&-24x&-20\\&& x^4&+3x^3&-10x^2&\\\cline{3-6}&&&3x^3&+11x^2&-24x\\&&&3x^3&+9x^2 &-30x\\\cline{4-7}&&&&2x^2&+6x&-20\\&&&&2x^2&+6x&-20\\\cline{5-7}&&&&0&0&0\\\end{tabular}\end{array}}$

• Remainder Obtained will be Other Two Zeroes of the Polynomial.

$\underline{\bigstar\:\:\textsf{By Splitting Middle Term :}}$

$:\implies\sf$ x² + 3x + 2 = 0

$:\implies\sf$ x² + (2 + 1)x + 2 = 0

$:\implies\sf$ x² + 2x + x + 2 = 0

$:\implies\sf$ x(x + 2) + 1(x + 2) = 0

$:\implies\sf$ (x + 1)(x + 2) = 0

$:\implies\sf$ x = - 1⠀or,⠀x = - 2

⠀

$\therefore$ Hence, All 4 Zeroes of Given Polynomial are 2, - 5, - 1 and, - 2.