Question

divide px by qx where px =x+3x2-1and qx = 1+x​

Answer 1

Step-by-step explanation:

A polynomial p(x) is defined as

⇒p(x)=g(x)q(x)+r(x)

where g(x)= divisor ; q(x)= quotient and r(x)= remainder

∴ p(x) can be found by multiplying g(x) with q(x) & adding r(x) to the product.

(i).g(x)=(x−2); q(x)=x

2

−x+1; r(x)=4

∴p(x)=(x−2)[x

2

−x+1]+4

=x

3

−x

2

+x−2x

2

+2x−2+4

=x

3

−3x

2

+3x+2

(ii).g(x)=(x+3); q(x)=2x

2

+x+5; r(x)=3x+1

∴p(x)=(x+3)[2x

2

+x+5]+(3x+1)

=2x

3

+x

2

+5x+6x

2

+3x+15+3x+1

=2x

3

+7x

2

+11x+16

(iii).g(x)=(2x+1); q(x)=x

3

+3x

2

−x+1; r(x)=0

∴p(x)=(2x+1)[x

3

+3x

2

−x+1]+(0)

=2x

4

+6x

3

−2x

2

+2x+x

3

+3x

2

−x+1

=2x

4

+7x

3

+x

2

+x+1

(iv).g(x)=(x−1); q(x)=x

3

−x

2

−x−1; r(x)=2x−4

∴p(x)=(x−1)[x

3

−x

2

−x−1]+(2x−4)

=x

4

−x

3

−x

2

−x−x

3

+x

2

+x+1+2x−4

=x

4

−2x

3

+2x−3

(v).g(x)=(x

2

+2x+1); q(x)=x

4

−2x

2

+5x−7; r(x)=4x+12

∴p(x)=(x

2

+2x+1)[x

4

−2x

2

+5x−7]+(4x+12)

=x

6

−2x

4

+5x

3

−7x

2

+2x

5

+4x

3

+10x

2

−14x+x

4

−2x

2

+5x−7+4x+12

=x

6

−x

4

+x

3

+x

2

+2x

5

−5x+5

=x

6

+2x

5

−x

4

+x

3

+x

2

−5x+5

Hence, solve.

Answer 2

$divide$

$= \frac{px}{qx}$

$px = x + 3x {}^{2} - 1$

$qx = 1 + x$

$now$

$\frac{x + 3 {x}^{2} - 1 }{1 + x}$

$\frac{x + 3 \times x \times x - 1}{1 + x}$

$x + 3 \times x - 1$

$x + 3x - 1$

$3 {x}^{2} - 1$

$2 {x}^{2}$