Question

Divide the polynomial p (x) and find quotient and remainder p (x) = 2x^4-3x^3-3x^2 -6x -2 ,g (x) =x^2 -2

Answer 1

Question :-

Divide the polynomial p(x) and find the quotient and remainder.

• $p(x) = {2x}^{4} - {3x}^{3} - {3x}^{2} - 6x - 2$
• $g(x) = {x}^{2} - 2$

Solution :-

$(\star \underline{\: Refer \: attachment \: for \: division.})$

• Remainder = -12x
• Quotient = 2x²- 3x + 1

Answer 2

Answer:

• p(x) = 2x⁴ – 3x³ – 3x² – 6x – 2
• g(x) = x² – 2

$\underline{\bigstar\:\:\textsf{According to the Question :}}$

$\dashrightarrow\tt\:\:p(x)=g(x) \times q(x) + r(x)\\\\\\\dashrightarrow\tt\:\:(2x^4-3x^3-3x^2-6x-2)=[(x^2-2)\times q(x)]+r(x)\\\\\\\dashrightarrow\tt\:\:\underline{\boxed{\tt{q(x)=2x^2-3x-1\quad and \quad r(x)=-\:12x }}}$

$\rule{160}{1}$

$\boxed{\begin{array}{l | c | r}(\sf x^2-2) &\sf 2x^4-3x^3-3x^2-6x-2&\sf2x^2-3x-1\\ &\sf2x^4\qquad\:\:-4x^2\qquad\qquad\\ &( - )\:\:\qquad(+)\quad\qquad\qquad\\&\rule{90}{0.8}\quad\:\:\:\\&\sf\:\:-3x^3+x^2-6x\\ &\sf-\:3x^3\qquad+6x\\ &( + )\quad\quad(-)\\&\quad\rule{85}{0.8}\\&\qquad\qquad\sf x^2-12x-2\\ &\sf\qquad\qquad\sf x^2\qquad\quad\!+2\\&\qquad\quad(-)\qquad(-)\\&\qquad\quad\rule{65}{0.8}\\ &\sf\qquad\qquad-\:12x\end{array}}$