Question

using remainder theorem find the remainder when x cube + x square - 2 x minus 3 is divided by 2 X + 3​

Answer 1

$\huge\mathbb{A~N~S~W~E~R}$

Let we,

$p(x) = 2x + 3 \\ \\ p(0) = 2x + 3 \\ \\ 0 = 2x + 3 \\ \\ - 3 = 2x \\ \\ x = \frac{ - 3}{2}$

Use value of x in fiven polynomial.

$= {x}^{3} + {x}^{2} - 2x + 3 \\ \\ = {( \frac{- 2}{3}) }^{3} + {( \frac{ - 2}{3} )}^{2} - 2( \frac{ - 2}{3} ) + 3 \\ \\ = \frac{ - 8}{27} + \frac{4}{9} + \frac{4}{3} + 3 \\ \\ = \frac{ - 8 + 12 + 36 + 81}{27} \\ \\ = \frac{121}{27}$

${\huge\color{Red}{Answer~is~\frac{121}{27}~ . }}$

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Answer 2

The remainder is -9/8.

Step-by-step explanation:

The given polynomial is

$f(x)=x^3+x^2-2x-3$

Remainder theorem: If a polynomial is divided by P(x) is divided by (x-c), then remainder is P(c).

We need to find the remainder when the given polynomial is divide by (2x+3).

$2x+3=0\Rightarrow x=-\dfrac{3}{2}$

Using remainder theorem the remainder is f(-3/2).

Substitute x=-3/2 in the given function.

$f(-\dfrac{3}{2})=(-\dfrac{3}{2})^3+(-\dfrac{3}{2})^2-2(-\dfrac{3}{2})-3$

$f(-\dfrac{3}{2})=-\dfrac{27}{8}+\dfrac{9}{4}+3-3$

$f(-\dfrac{3}{2})=\dfrac{-27+19}{8}$

$f(-\dfrac{3}{2})=-\dfrac{9}{8}$

Therefore, the remainder is -9/8.

#Learn more

The polynomials p(x)=ax³+4x²+3x-4 and q(x)=x³-4x+a leaves the same remainder when divided by (x-3) find the remainder when p(x) is divided by (x-2).

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