(i) p(x) = $x^3 – 3x^2$ + 5x – 3, g(x) = $x^2$ – 2
(ii) p(x) = $x^4 – 3x^2$ + 4x + 5, g(x) = $x^2$ + 1 – x
(iii) p(x) = $x^4$ – 5x + 6, g(x) = 2 –$ x^2$
Q1.(i) p(x) = $x^3 – 3x^2$ + 5x – 3, g(x) = $x^2$ – 2
Given,
Dividend = p(x) = $x^3-3x^2$+5x–3
Divisor = g(x) = $x^2$ – 2
Then, divide the polynomial p(x) by g(x)
Therefore,
Quotient = x–3
Remainder = 7x–9
(ii) p(x) = $x^4 – 3x^2$ + 4x + 5, g(x) = $x^2$ + 1 – x
Given,
Dividend = p(x) = $x^4 +0x^3– 3x^2$ + 4x + 5
Divisor = g(x) = $x^2$ + 1 – x
Therefore,
Quotient = $x^2$ + x–3
Remainder = 8
(iii) p(x) = $x^4$ – 5x + 6, g(x) = 2 –$ x^2$
Given,
Dividend = p(x) =$x^4$ – 5x + 6 = $x^4+0x^3+0x^2$–5x+6
Divisor = g(x) = $2–x^2 = –x^2+2$
Therefore,
Quotient = $-x^2-2$
Remainder = -5x + 10
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) $t^2 – 3, 2t^4 + 3t^3 – 2t^2 – 9t – 12$
(ii) $x^2 + 3x + 1, 3x^4 + 5x^3 – 7x^2 + 2x + 2$
(iii) $x^3 – 3x + 1, x^5 – 4x^3 + x^2 + 3x + 1$
(i) $t^2 – 3, 2t^4 + 3t^3 – 2t^2 – 9t – 12$
Given,
First polynomial = $t^2+0t-3$
Second polynomial = $2t^4 +3t^3-2t^2 -9t-12$
Now, divide the second polynomial by the first polynomial.
Since the remainder is 0
Therefore, $t^2−3$ is a factor of $2t^4+3t^3−2t^2−9t−12$
(ii) $x^2 + 3x + 1, 3x^4 + 5x^3 – 7x^2 + 2x + 2$
Given,
First polynomial = $4x^2+3x+1$
Second polynomial = $3x^4+5x^3-7x^2+2x+2$
Now, divide the second polynomial by the first polynomial.
Since the remainder is 0
Therefore, $x^2+ 3x + 1$ is a factor of $3x^4+ 5x^3−7x^2+ 2x + 2$
(iii) $x^3 – 3x + 1, x^5 – 4x^3 + x^2 + 3x + 1$
Given,
First polynomial = $x^3 – 3x + 1$
Second polynomial = $x^5 – 4x^3 + x2 + 3x + 1$
Now, divide the second polynomial by the first polynomial.
Since the remainder is $\neq$ 0
Therefore, $x^3 – 3x + 1$ is not a factor of $x^5 – 4x^3 + x^2 + 3x + 1$
Since this is a polynomial equation of degree 4, there will be a total of 4 roots.
Then, given two zeroes are $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$
(x-$\sqrt{\dfrac{5}{3}}$) (x+$\sqrt{\dfrac{5}{3}}$) = $x^2 - \dfrac{5}{3}$
$x^2 - \dfrac{5}{3}$ = 0
$3x^2 - 5 = 0$, is a factor of given polynomial
Now, divide the given polynomial $3x^2 - 5$
Therefore, $3x^4 +6x^3 −2x^2 −10x–5 = (3x^2 –5)(x^2+2x+1)$
Now, on further factorising $(x^2+2x+1)$, we get
$x^2+2x+1 = x^2+x+x+1 = 0$
x(x+1)+1(x+1) = 0
(x+1)(x+1) = 0
So, its zeroes are given by: x= −1 and x = −1
Therefore, all four zeroes of the given polynomial equation are
$\sqrt{\dfrac{5}{3}}$ , $-\sqrt{\dfrac{5}{3}}$ , −1 and −1
Given,
Dividend, p(x) = $x^3-3x^2+x+2$
Quotient = x-2
Remainder = –2x+4
g(x) =?
Dividend = Divisor × Quotient + Remainder
$x^3-3x^2+x+2$ = g(x) $\times$ (x-2) + (-2x+4)
$x^3-3x^2+x+2-(-2x+4)$ = g(x) $\times$ (x-2)
Therefore, g(x) $\times$ (x-2) = $x^3-3x^2+3x-2$
Now, for finding g(x), we will divide $x^3-3x^2+3x-2$ with (x-2)
Therefore, g(x)=$(x^2−x+1)$
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then, we can find the value of quotient q(x) and remainder r(x) with the help of below-given formula.
| Dividend = Divisor × Quotient + Remainder |
p(x) = g(x) $\times$ q(x) + r(x)
Where r(x) = 0 or degree of r(x)< degree of g(x)
Now, let us prove the three given cases, as per the division algorithm, by taking examples for each.
(i) deg p(x) = deg q(x)
The degree of dividend is equal to the degree of the quotient only when the divisor is a constant term.
Let us take an example:
p(x) = $3x^2+3x+3$ is a polynomial to be divided by g(x) = 3
So, $\dfrac{3x^2+3x+3}{3} = x^2+x+1 = q(x)$
Thus, you can see the degree of quotient q(x) = 2, which is also equal to the degree of dividend p(x).
Hence, the division algorithm is satisfied here.
(ii) deg q(x) = deg r(x)
Let us take an example:
p(x) = $x^2 + 3$ is a polynomial to be divided by g(x) = x – 1
So, $x^2 + 3$ = (x – 1) $\times$ (x) + (x + 3)
Hence, quotient q(x) = x
Also, remainder r(x) = x + 3
Thus, you can see the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).
Hence, the division algorithm is satisfied here.
(iii) deg r(x) = 0
The degree of remainder is 0 only when the remainder left after the division algorithm is constant.
Let us take an example:
p(x) = $x^2 + 1$ is a polynomial to be divided by g(x) = x.
So, $x^2 + 1$ = (x) $\times$ (x) + 1
Hence, quotient q(x) = x
And, remainder r(x) = 1
Clearly, the degree of remainder here is 0.
Hence, the division algorithm is satisfied here.