EXERCISE-4

Example-8: Divide 2x² + 3x + 1 by x + 2.

Solution :

Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.

So, here the quotient is 2x – 1 and the remainder is 3.

Also,

(2x – 1) (x + 2) + 3 = 2x² + 4x – x – 2 + 3 = 2x² + 3x + 1

i.e., 2x² + 3x + 1 = (x + 2) (2x – 1) + 3

∴ Dividend = Divisor × Quotient + Remainder

Example-9: Divide 3x³ + x² + 2x + 5 by 1 + 2x + x².

Solution :

We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Arranging the terms in this order is termed as writing the polynomials in its standard form.

Step 1 :

To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e., 3x³) by the highest degree term of the divisor (i.e., x²). This is 3x. Then carry out the division process. What remains is – 5x² – x + 5.

Step 2 :

Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e., – 5x²) by the highest degree term of the divisor (i.e., x²). This gives – 5. carry out the division process with– 5x² – x + 5.

Step 3 :

What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x² + 2x + 1. So, we cannot continue the division any further. So, the quotient is 3x – 5 and the remainder is 9x + 10.

(x² + 2x + 1) × (3x – 5) + (9x + 10) = (3x ³ + 6x ² + 3x – 5x ² – 10x – 5 + 9x + 10) = 3x ³ + x² + 2x + 5

we see that

Dividend = Divisor × Quotient + Remainder

Example-10: Divide 3x² – x³ – 3x + 5 by x – 1 – x², and verify the division algorithm.

Solution :

Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees.

So, dividend = – x³ + 3x²– 3x + 5 and divisor = – x²+ x – 1.

We stop here since degree of the remainder is less than the degree of (–x² + x – 1) the divisor.

So, quotient = x – 2, remainder = 3.

Now, Dividend = Divisor × Quotient + Remainder

= (–x² + x – 1) (x – 2) + 3

= – x³+ x² – x + 2x² – 2x + 2 + 3

= – x³ + 3x² – 3x + 5

In this way, the division algorithm is verified.

Example-11: Find all the zeroes of 2x⁴ – 3x³ – 3x² + 6x – 2, if you know that two of its zeroes are `sqrt2` , – `sqrt2`.

Solution :

Since two of the zeroes are `sqrt2` , – `sqrt2`, therefore we can divide by (x – `sqrt2`) (x + `sqrt2` ) = x² – 2.

So, 2x⁴ – 3x³ – 3x² + 6x – 2 = (x² – 2) (2x² – 3x + 1).

Now, by splitting –3x, we factorize 2x² – 3x + 1 as

2x² – 2x – x+ 1= 2x(x – 1)– 1(x – 1)= (2x – 1) (x – 1).

So, its zeroes are given by x =`1/2`and x = 1.

∴ The zeroes of the given polynomial are `sqrt2` , – `sqrt2`, 1 and `1/2`.

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² –2

Solution :

p(x) = x³ – 3x² + 5x – 3

g(x) = x² –2

quotient =x-3

remainder = 7x-9

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

Solution :

p(x) = x⁴ +0x³– 3x² + 4x + 5

g(x) = x² – x + 1

quotient =x² + x - 3

remainder = 8

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 –x²

Solution :

p(x) = x⁴+0x³ + 0x² – 5x + 6

g(x) = –x²+0x+ 2

quotient =-x² - 2

remainder = -5x + 10

2. Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :

(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12

Solution :

Remainder is zero. So t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

Solution :

Remainder is zero. So x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution :

Remainder is two. So x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1

3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are `(sqrt(5/3),-sqrt(5/3)`.

Solution :

(`sqrt(5/3)`,`-sqrt(5/3)` are zeroes of p(x).So (x`-sqrt(5/3)`)(x+`sqrt(5/3)`)=(x²`-5/3`) be a factor to p(x).

Remainder is zero and

quotient=x²+2x+1

x²+2x+1 = (x+1)²

(x+1)²=0 ⇒ x+1=0

∴ x = -1.

So the zeroes of p(x) are -1,-sqrt(5/3) and sqrt(5/3).

4. On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x).

Solution :

p(x)=x³ – 3x² + x + 2

q(x)= x - 2

r(x) = -2x+4

Division algorithm for polynomials is p (x) = q (x) . g (x) + r (x)

p(x)-r(x)=q (x) . g (x)

x³ – 3x² + x + 2 - (-2x+4)=(x-2).g(x)

g(x) = `(x<sup>3</sup> – 3x<sup>2</sup> + 3x - 2 )/(x-2)`

∴g(x) = x²-x+1

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

Solution :

Division algorithm for polynomials is p (x) = g (x) . q (x) + r (x)

For deg p(x)= deg q(x), deg g(x) must be 0.So g(x) is a constant polynomial.

Let's we take

g(x)=2

r(x)=1

q(x)=3x² -2x+3

p(x)=2(3x² -2x+3)+1

∴ p(x)=6x² -4x+7

(ii) deg q(x) = deg r(x)

Solution :

Division algorithm for polynomials is p (x) = g (x) . q (x) + r (x)

We know that deg r(x) < g(x) For deg q(x) = deg r(x) it must be deg q(x) < g(x).

Let's we take

g(x)=x²+2x+1

q(x)=x+2

r(x)=x-1, then

p(x)=(x²+2x+1)(x+2)+(x-1)

= x³+2x²+2x²+4x+x+2+x-1

=x³+4x²+6x+1

∴p(x)=x³+4x²+6x+1

(iii) deg r(x) = 0

Solution :

Division algorithm for polynomials is p (x) = g (x) . q (x) + r (x)

deg r(x) = 0 means r(x) is a constant polynomial. Let's we take

r(x)=3

g(x)=2

q(x)=(x+1)

then p(x)= 2(x+1)+3=2x+5

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