Question

Find all other zeroes of the polynomial p(x)=2x^4-7x^3+19x^2-14x+30 if two of zeroes are √2 and -√2
Please answer step by step​

Answer 1

$\sf\large\underline{Correct\: Question:}$

If polynomial 2x^4-3x^3-3x^2+6x-2 are The two zeroes of √2 and -√2. Find the other polynomial of zeroes.

$\sf\large\underline{Given:}$

• $\rm{P(x)=2x^4-3x^3-3x^2+6x-2}$
• $\rm{The\: zeroes\:are\:\sqrt{2},\:-\sqrt{2}}$

$\sf\large\underline{To\: Find:}$

• $\rm{The\: other\:zeroes\:of\:P}$

$\sf\large\underline{Solution:}$

• At 1st find out the factor of polynomial with the help of given zeroes than divide, we know that given polynomial is exactly divisible by the given factors

$\rm{\implies (x+\sqrt{2})(x-\sqrt{2})}$

$\rm{\implies x^2-2}$

• Now notice on refers attachment here
• By solving we get here]

$\rm{\implies P(x)=(x^2-2)(2x^2-3x+1)}$

$\rm{\implies (x^2-2)(2x^2-2x-x+1)}$

$\rm{\implies (x^2-2)[2x(x-1)-1(x-1)]}$

$\rm{\implies (x^2-2)(2x-1)(x-1)}$

• Now find out the other zeroes here]

$\rm{\implies 2x-1=0}$

$\rm{\implies 2x=1=>x=\dfrac{1}{2}}$

$\rm{\implies x-1=0}$

$\rm{\implies x=1}$

$\sf\large{Hence,}$

$\rm{\implies The\: other\: zeroes\:are\:1,\:\dfrac{1}{2}}$

Answer 2

Correct Question:

Find all other zeroes of the polynomial $\rm p(x)=2x^4-3x^3-3x^2+6x-2$ if two of zeroes are $-\:\sqrt{2}$ and $\sqrt{2}$.

$\rule{110}{1}$

Answer:

• Polynomial : $\rm 2x^4-3x^3-3x^2+6x-2$
• Zeroes : $-\:\sqrt{2}$ and $\sqrt{2}$

If $-\:\sqrt{2}$ and $\sqrt{2}$ are zeroes of the Polynomial then, Polynomial will be Divisible by $\rm[x-(-\:\sqrt{2})]$ and $\rm[x-\sqrt{2}]$ and so to there Products.

⇢ $\rm[x-(-\:\sqrt{2})][x-\sqrt{2}]$

⇢ $\rm[x+\sqrt{2}][x-\sqrt{2}]$

⇢ $\rm x^2-2x$

• (x² - 2) will completely Divide Polynomial. So by Division.

$\rule{150}{2}$

$\boxed{\begin{array}\quad\begin{tabular}{m{3.5em}ccccc}&&\bf 2x^2&\bf- 3x&\bf+ 1\\\cline{1-7}\multicolumn{2}{l}{\sf x{^\sf2} - \sf2\big)}&2x^4&- 3x^3&- 3x^2&+ 6x&- 2\\&& - (2x^4&&- 4x^2)&\\\cline{3-6}&&&- 3x^3&+x^2&+6x\\&&&- (-3x^3& &+ 6x)\\\cline{4-7}&&&&x^2&&-2\\&&&&- (x^2&&-2)\\\cline{5-7}&&&&\times&&\times\\\end{tabular}\end{array}}$

• Remainder Obtained will be other two zeroes of the Polynomial.

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

$:\implies$ 2x² – 3x + 1 = 0

$:\implies$ 2x² – (2 + 1)x + 1 = 0

$:\implies$ 2x² – 2x – x + 1 = 0

$:\implies$ 2x(x – 1) – 1(x – 1) = 0

$:\implies$ (x – 1)(2x – 1) = 0

$:\implies$ x = 1⠀or,⠀x = ½

⠀

∴ Hence, All four Zeroes of the given polynomial are $\sf\sqrt{2},-\:\sqrt{2},1\:and\:{}^{1}\!/{}_{2}$