How do you divide (2x^3 - 11x^2 + 12x + 9)/(x-3)2x311x2+12x+9x3?

2 Answers
Jul 27, 2018

The remainder is 00 and the quotient is =2x^2-5x-3=2x25x3

Explanation:

Let's perform the synthetic division

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The remainder is 00 and the quotient is =2x^2-5x-3=2x25x3

(2x^3-11x^2+12x+9)/(x-3)=2x^2-5x-32x311x2+12x+9x3=2x25x3

Apply the remainder theorem

When a polynomial f(x)f(x) is divided by (x-c)(xc), we get

f(x)=(x-c)q(x)+rf(x)=(xc)q(x)+r

Let x=cx=c

Then,

f(c)=0+rf(c)=0+r

Here,

f(x)=2x^3-11x^2+12x+9f(x)=2x311x2+12x+9

Therefore,

f(3)=2*3^3-11*3^2+12*3+9f(3)=2331132+123+9

=54-99+36+9=5499+36+9

=0=0

The remainder is =0=0

Jul 27, 2018

(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)(2x311x2+12x+9)=(x3)(2x25x3)+(0)

Explanation:

(2x^3-11x^2+12x+9)div(x-3)(2x311x2+12x+9)÷(x3)

We can divide this polynomial by using synthetic division

We have , p(x)=2x^3-11x^2+12x+9 and "divisor :"x=3p(x)=2x311x2+12x+9anddivisor :x=3

We take ,coefficients of p(x) to 2,-11,12 ,9p(x)2,11,12,9

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We can see that , quotient polynomial :

q(x)=2x^2-5x-3 and"the Remainder"=0

Hence ,

(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)