What will be the remainder when 6x ^5+ 4x^4 -27x^ 3
- 7x² + 27x + 3/2 is divided by (2x^2 - 3)
^2

Remainder = (3145/8)x - 408

Step-by-step explanation:

We want to find the remainder when 6x^(5) + 4x⁴ - 27x³ - 7x² + 27x + 3/2 is divided by (2x² - 3)²

Let's expand (2x² - 3)² to give ;

(2x - 3)(2x - 3) = 4x² - 6x - 6x + 9 = 4x² - 12x + 9

So,we can divide now;

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

First of all, we'll divide the term with the highest power inside the long division symbol by the term with the highest power outside the division symbol. This will give;

3/2x³

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

We now subtract the new multiplied term beneath the original one from the original one to get;

3/2x³

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³+27x +3/2

We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;

(3/2)x³ + (11/2)x²

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

We now subtract the new multiplied term beneath the immediate one from the immediate one to get;

(3/2)x³ + (11/2)x²

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;

(3/2)x³ + (11/2)x² + (159/8)x

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

(159/2)x³-(477/2)x²+(1431/8)x

We now subtract the new multiplied term beneath the immediate one from the immediate one to get;

(3/2)x³ + (11/2)x² + (159/8)x

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

(159/2)x³-(477/2)x²+(1431/8)x

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

182x²-(1215/8)x + (3/2)

We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;

(3/2)x³+(11/2)x²+(159/8)x+(91/2)

4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2

6x^(5)-18x⁴-(27/2)x³

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

22x⁴+(27/2)x³-7x²+27x +3/2

22x⁴-66x³ + (99/2)x²

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

(159/2)x³-(113/2)x²+27x+(3/2)

(159/2)x³-(477/2)x²+(1431/8)x

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

182x²-(1215/8)x + (3/2)

182x²-545x + 819/2

We now subtract the new multiplied term beneath the immediate one from the immediate one to get;

182x² - (1215/8)x + (3/2) - 182x² + 545x - 819/2 = (3145/8)x - 408

Remainder = (3145/8)x - 408

it would be:

c.parallelogram.

.bjbjk.b,j.l.

step-by-step explanation: