Mathematics, 26.08.2020 21:01 aliay6857

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

Answers: 1

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Answer from: mine9226

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

Answer from: Quest

your answer is a

step-by-step explanation:

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Answer from: Quest

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

$ab^2+ac^2=bc^2$

$ab^2+ac^2=x^2$ )

triangle abd and cba are similar,

$\frac{ab}{bc}=\frac{bd}{ab}$

$ab^2=bc\cdot bd$

$25=bc\cdot bd$ (2)

triangle cad and cba are similar,

$\frac{ac}{bc}=\frac{cd}{ac}$

$ac^2=bc\cdot dc$

$7\times 7=bc\cdot dc$

$49=bc\cdot dc$ )

using (1), (2) and (3), we get

$25+49=x^2$

$74=x^2$

hence proved.

therefore option 2 is correct.

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What will be the remainder when 6x ^5+ 4x^4 -27x^ 3
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