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
