Answer the following and justify:
(i) Can x2 – 1 be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5?
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No, x2 – 1 cannot be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5.
Justification:
When a degree 6 polynomial is divided by degree 5 polynomial,
The quotient will be of degree 1.
Assume that (x2 – 1) divides the degree 6 polynomial with and the quotient obtained is degree 5 polynomial (1)
According to our assumption,
(degree 6 polynomial) = (x2 – 1)(degree 5 polynomial) + r(x) [ Since, (a = bq + r)]
= (degree 7 polynomial) + r(x) [ Since, (x2 term × x5 term = x7 term)]
= (degree 7 polynomial)
From the above equation, it is clear that, our assumption is contradicted.
x2 – 1 cannot be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5
Hence Proved.
(ii) What will the quotient and remainder be on division of ax2 + bx + c by px3 + qx2 + rx + s, p ≠ 0?
Degree of the polynomial px3 + qx2 + rx + s is 3
Degree of the polynomial ax2 + bx + c is 2
Here, degree of px3 + qx2 + rx + s is greater than degree of the ax2 + bx + c
Therefore, the quotient would be zero,
And the remainder would be the dividend = ax2 + bx + c.
(iii) If on division of a polynomial p (x) by a polynomial g (x), the quotient is zero, what is the relation between the degrees of p (x) and g (x)?
We know that,
p(x)= g(x) × q(x)+r(x)
According to the question,
q(x) =0
When q(x)=0, then r(x) is also = 0
So, now when we divide p(x) by g(x),
Then p(x) should be equal to zero
Hence, the relation between the degrees of p (x) and g (x) is the degree p(x)<degree g(x)
(iv) If on division of a non-zero polynomial p (x) by a polynomial g (x), the remainder is zero, what is the relation between the degrees of p (x) and g (x)?
In order to divide p(x) by g(x)
We know that,
Degree of p(x) > degree of g(x)
or
Degree of p(x)= degree of g(x)
Therefore, we can say that,
The relation between the degrees of p (x) and g (x) is degree of p(x) > degree of g(x)
(v) Can the quadratic polynomial x2 + kx + k have equal zeroes for some odd integer k > 1?
A Quadratic Equation will have equal roots if it satisfies the condition:
b² – 4ac = 0
Given equation is x² + kx + k = 0
a = 1, b = k, x = k
Substituting in the equation we get,
k² – 4 ( 1 ) ( k ) = 0
k² – 4k = 0
k ( k – 4 ) = 0
k = 0 , k = 4
But in the question, it is given that k is greater than 1.
Hence the value of k is 4 if the equation has common roots.
Hence if the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.