Page No 52:
Answer:
Page No 52:
Question 2:
Answer:
Sum of zeroes =
−4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes =
(−4)(−3)=121=constant termcoefficient of x2
Page No 52:
Question 3:
Sum of zeroes =
−4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes =
(−4)(−3)=121=constant termcoefficient of x2
Answer:
Sum of zeroes =
Product of zeroes =
Page No 52:
Question 4:
Sum of zeroes =
Product of zeroes =
Answer:
Let f(x) = 2x2 – x – 6
Hence, all the zeroes of the polynomial f(x) are .
Now,
Hence, the relationship between the zeros and the coefficients is verified.
Page No 52:
Question 5:
Let f(x) = 2x2 – x – 6
Hence, all the zeroes of the polynomial f(x) are .
Now,
Hence, the relationship between the zeros and the coefficients is verified.
Answer:
Page No 52:
Question 6:
Answer:
Page No 52:
Question 7:
Answer:
Page No 52:
Question 8:
Answer:
Sum of zeroes =
−4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes =
(−4)(−3)=121=constant termcoefficient of x2
Page No 52:
Question 9:
Sum of zeroes =
−4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes =
(−4)(−3)=121=constant termcoefficient of x2
Answer:
Sum of zeroes =
−4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes =
(−4)(−3)=121=constant termcoefficient of x2
Page No 52:
Question 10:
Sum of zeroes =
−4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes =
(−4)(−3)=121=constant termcoefficient of x2
Answer:
Let f(x) = 3x2 – x – 4
Hence, all the zeroes of the polynomial f(x) are .
Now,
Hence, the relationship between the zeros and the coefficients is verified.
Page No 52:
Question 11:
Let f(x) = 3x2 – x – 4
Hence, all the zeroes of the polynomial f(x) are .
Now,
Hence, the relationship between the zeros and the coefficients is verified.
Answer:
Let f(x) = 5x2 + 10x
Hence, all the zeroes of the polynomial f(x) are .
Now,
Hence, the relationship between the zeros and the coefficients is verified.
Page No 52:
Question 12:
Let f(x) = 5x2 + 10x
Hence, all the zeroes of the polynomial f(x) are .
Now,
Hence, the relationship between the zeros and the coefficients is verified.
Answer:
Page No 52:
Question 13:
Answer:
Let α and β be the zeroes of the polynomial .
Now, using (1)
Hence, the value of k is 2.
Page No 52:
Question 14:
Let α and β be the zeroes of the polynomial .
Now, using (1)
Hence, the value of k is 2.
Answer:
Page No 52:
Question 15:
Answer:
Page No 52:
Question 16:
Answer:
Page No 52:
Question 17:
Answer:
Page No 52:
Question 18:
Answer:
Page No 52:
Question 19:
Answer:
Given: is a factor of
So, we have
Now, It will satisfy the above polynomial.
Therefore, we will get
Page No 53:
Question 20:
Given: is a factor of
So, we have
Now, It will satisfy the above polynomial.
Therefore, we will get
Answer:
Given:
Since, is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
Since, is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
From (1) and (2), we get
Page No 59:
Question 18:
Given:
Since, is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
Since, is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
From (1) and (2), we get
Answer:
Page No 60:
Question 23:
Answer:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Page No 61:
Question 24:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Answer:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Page No 61:
Question 25:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Answer:
By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes =
Now, Product of zeros =
Page No 63:
Question 1:
By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes =
Now, Product of zeros =
Answer:
Page No 63:
Question 2:
Answer:
Page No 63:
Question 3:
Answer:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
...(1)
Let
Substituting the values in (1), we get
Page No 63:
Question 4:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
...(1)
Let
Substituting the values in (1), we get
Answer:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
...(1)
Let
Substituting the values in (1), we get
Page No 63:
Question 5:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
...(1)
Let
Substituting the values in (1), we get
Answer:
We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x3 −(Sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x − Product of zeroes
Therefore, the required polynomial is
Page No 63:
Question 6:
We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x3 −(Sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x − Product of zeroes
Therefore, the required polynomial is
Answer:
Quotient
Remainder
Page No 63:
Question 7:
Quotient
Remainder
Answer:
Quotient
Remainder
Page No 63:
Question 8:
Quotient
Remainder
Answer:
We can write
and
Quotient
Remainder
Page No 63:
Question 9:
We can write
and
Quotient
Remainder
Answer:
Let and
Quotient
Remainder
Since, the remainder is 0.
Hence, is a factor of
Page No 63:
Question 10:
Let and
Quotient
Remainder
Since, the remainder is 0.
Hence, is a factor of
Answer:
Let f(x) = x4 + 2x3 + 8x2 + 12x + 18
It is given that when f(x) is divisible by x2 + 5, the remainder comes out to be px + q.
On division, we get the quotient x2 + 2x + 3 and the remainder 2x + 3.
Since, the remainder comes out to be px + q.
Therefore, p = 2 and q = 3.
Hence, the values of p and q are 2 and 3 respectively.
Page No 63:
Question 11:
Let f(x) = x4 + 2x3 + 8x2 + 12x + 18
It is given that when f(x) is divisible by x2 + 5, the remainder comes out to be px + q.
On division, we get the quotient x2 + 2x + 3 and the remainder 2x + 3.
Since, the remainder comes out to be px + q.
Therefore, p = 2 and q = 3.
Hence, the values of p and q are 2 and 3 respectively.
Answer:
By using division rule, we have
Divided = Quotient × Divisor + Remainder
Page No 63:
Question 12:
By using division rule, we have
Divided = Quotient × Divisor + Remainder
Answer:
We can write and
Quotient =
Remainder =
By using division rule, we have
Divided = Quotient × Divisor + Remainder
Page No 63:
Question 13:
We can write and
Quotient =
Remainder =
By using division rule, we have
Divided = Quotient × Divisor + Remainder
Answer:
Page No 63:
Question 14:
Answer:
Page No 63:
Question 15:
Answer:
Page No 63:
Question 16:
Answer:
Let f(x) = 2x4 – 5x3 – 11x2 + 20x + 12
It is given that 2 and –2 are two zeroes of f(x)
Thus, f(x) is completely divisible by (x + 2) and (x – 2).
Therefore, one factor of f(x) is (x2 – 4).
We get another factor of f(x) by dividing it with (x2 – 4).
On division, we get the quotient 2x2 – 5x – 3.
Hence, all the zeroes of the polynomial f(x) are
Page No 64:
Question 17:
Let f(x) = 2x4 – 5x3 – 11x2 + 20x + 12
It is given that 2 and –2 are two zeroes of f(x)
Thus, f(x) is completely divisible by (x + 2) and (x – 2).
Therefore, one factor of f(x) is (x2 – 4).
We get another factor of f(x) by dividing it with (x2 – 4).
On division, we get the quotient 2x2 – 5x – 3.
Hence, all the zeroes of the polynomial f(x) are
Answer:
Page No 64:
Question 19:
Answer:
Let f(x) = x4 + x3 – 14x2 – 2x + 24
It is given that are two zeroes of f(x)
Thus, f(x) is completely divisible by (x + ) and (x – ).
Therefore, one factor of f(x) is (x2 – 2).
We get another factor of f(x) by dividing it with (x2 – 2).
On division, we get the quotient x2 + x – 12.
Hence, all the zeroes of the polynomial f(x) are
Page No 64:
Question 20:
Let f(x) = x4 + x3 – 14x2 – 2x + 24
It is given that are two zeroes of f(x)
Thus, f(x) is completely divisible by (x + ) and (x – ).
Therefore, one factor of f(x) is (x2 – 2).
We get another factor of f(x) by dividing it with (x2 – 2).
On division, we get the quotient x2 + x – 12.
Hence, all the zeroes of the polynomial f(x) are
Answer:
Let f(x) = 2x4 – 13x3 + 19x2 + 7x – 3
It is given that are two zeroes of f(x)
Thus, f(x) is completely divisible by (x ) and (x – ).
Therefore, one factor of f(x) is
one factor of f(x) is (x2 – 4x + 1)
We get another factor of f(x) by dividing it with (x2 – 4x + 1).
On division, we get the quotient 2x2 – 5x – 3.
Hence, all the zeroes of the polynomial f(x) are
Page No 64:
Question 21:
Let f(x) = 2x4 – 13x3 + 19x2 + 7x – 3
It is given that are two zeroes of f(x)
Thus, f(x) is completely divisible by (x ) and (x – ).
Therefore, one factor of f(x) is
one factor of f(x) is (x2 – 4x + 1)
We get another factor of f(x) by dividing it with (x2 – 4x + 1).
On division, we get the quotient 2x2 – 5x – 3.
Hence, all the zeroes of the polynomial f(x) are
Answer:
Let f(x) = 3x3 + 16x2 + 15x – 18
It is given that one of its zeroes is .
Therefore, one factor of f(x) is (x – ).
We get another factor of f(x) by dividing it with (x – ).
On division, we get the quotient 3x2 + 18x + 27.
Hence, other zero of the polynomial f(x) is –3.
Page No 64:
Question 22:
Let f(x) = 3x3 + 16x2 + 15x – 18
It is given that one of its zeroes is .
Therefore, one factor of f(x) is (x – ).
We get another factor of f(x) by dividing it with (x – ).
On division, we get the quotient 3x2 + 18x + 27.
Hence, other zero of the polynomial f(x) is –3.
Answer:
Let f(x) = 2x4 – 3x3 – 3x2 + 6x – 2
It is given that 1 and are two zeroes of f(x).
Thus, f(x) is completely divisible by (x – 1) and (x – ).
Therefore, one factor of f(x) is
one factor of f(x) is
We get another factor of f(x) by dividing it with .
On division, we get the quotient 2x2 – 4.
Hence, all the zeroes of the polynomial f(x) are
Page No 64:
Question 23:
Let f(x) = 2x4 – 3x3 – 3x2 + 6x – 2
It is given that 1 and are two zeroes of f(x).
Thus, f(x) is completely divisible by (x – 1) and (x – ).
Therefore, one factor of f(x) is
one factor of f(x) is
We get another factor of f(x) by dividing it with .
On division, we get the quotient 2x2 – 4.
Hence, all the zeroes of the polynomial f(x) are
Answer:
Page No 65:
Question 1:
Answer:
Let the other zeroes of be a.
By using the relationship between the zeroes of the quadratic ploynomial.
We have, Sum of zeroes =
Hence, the other zeroes of is .
Page No 65:
Question 2:
Let the other zeroes of be a.
By using the relationship between the zeroes of the quadratic ploynomial.
We have, Sum of zeroes =
Hence, the other zeroes of is .
Answer:
By adding and subtracting
px, we get
So, the zeros of
f(
x) are −(
p + 1) and
p.
Page No 65:
Question 3:
By adding and subtracting
px, we get
So, the zeros of
f(
x) are −(
p + 1) and
p.
Answer:
By adding and subtracting
mx, we get
So, the zeros of
f(
x) are −
m and
m + 3.
Page No 66:
Question 4:
By adding and subtracting
mx, we get
So, the zeros of
f(
x) are −
m and
m + 3.
Answer:
If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as
x2 − (α + β)x + αβ .....(1)
Substituting the values in (1), we get
x2 − 6x + 4
Page No 66:
Question 5:
If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as
x2 − (α + β)x + αβ .....(1)
Substituting the values in (1), we get
x2 − 6x + 4
Answer:
Given: x = 2 is one zero of the quadratic polynomial kx2 + 3x + k
Therefore, It will satisfy the above polynomial.
Now, we have
Page No 66:
Question 6:
Given: x = 2 is one zero of the quadratic polynomial kx2 + 3x + k
Therefore, It will satisfy the above polynomial.
Now, we have
Answer:
Given: x = 3 is one zero of the polynomial 2x2 + x + k
Therefore, It will satisfy the above polynomial.
Now, we have
Page No 66:
Question 7:
Given: x = 3 is one zero of the polynomial 2x2 + x + k
Therefore, It will satisfy the above polynomial.
Now, we have
Answer:
Given: x = −4 is one zero of the polynomial x2 − x −(2k + 2)
Therefore, It will satisfy the above polynomial.
Now, we have
Page No 66:
Question 8:
Given: x = −4 is one zero of the polynomial x2 − x −(2k + 2)
Therefore, It will satisfy the above polynomial.
Now, we have
Answer:
Given: x = 1 is one zero of the polynomial ax2 − 3(a − 1) x − 1
Therefore, It will satisfy the above polynomial.
Now, we have
Page No 66:
Question 9:
Given: x = 1 is one zero of the polynomial ax2 − 3(a − 1) x − 1
Therefore, It will satisfy the above polynomial.
Now, we have
Answer:
Given: x = −2 is one zero of the polynomial 3x2 + 4x + 2k
Therefore, It will satisfy the above polynomial.
Now, we have
Page No 66:
Question 10:
Given: x = −2 is one zero of the polynomial 3x2 + 4x + 2k
Therefore, It will satisfy the above polynomial.
Now, we have
Answer:
So, the zeros of
f(
x) are 3 and −2.
Page No 66:
Question 11:
So, the zeros of
f(
x) are 3 and −2.
Answer:
By using the relationship between the zeros of the quadratic ploynomial.
We have
Sum of zeroes =
Page No 66:
Question 12:
By using the relationship between the zeros of the quadratic ploynomial.
We have
Sum of zeroes =
Answer:
By using the relationship between the zeros of the quadratic ploynomial.
We have
Product of zeroes =
Page No 66:
Question 13:
By using the relationship between the zeros of the quadratic ploynomial.
We have
Product of zeroes =
Answer:
Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10
We have
Since, (x + a) is a factor of 2x2 + 2ax + 5x + 10
Hence, It will satisfy the above polynomial
Page No 66:
Question 14:
Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10
We have
Since, (x + a) is a factor of 2x2 + 2ax + 5x + 10
Hence, It will satisfy the above polynomial
Answer:
By using the relationship between the zeroes of the cubic ploynomial.
We have
Sum of zeroes =
Page No 66:
Question 15:
By using the relationship between the zeroes of the cubic ploynomial.
We have
Sum of zeroes =
Answer:
Equating x2 − x to 0 to find the zeros, we will get
Since, x3 + x2 − ax + b is divisible by x2 − x.
Hence, the zeros of x2 − x will satisfy x3 + x2 − ax + b
and
Page No 66:
Question 16:
Equating x2 − x to 0 to find the zeros, we will get
Since, x3 + x2 − ax + b is divisible by x2 − x.
Hence, the zeros of x2 − x will satisfy x3 + x2 − ax + b
and
Answer:
By using the relationship between the zeros of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Page No 66:
Question 17:
By using the relationship between the zeros of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Answer:
“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that
f(x) = g(x) × q(x) + r(x),
where
r(
x) = 0 or degree of
r(
x) < degree of
g(
x).
Page No 66:
Question 18:
“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that
f(x) = g(x) × q(x) + r(x),
where
r(
x) = 0 or degree of
r(
x) < degree of
g(
x).
Answer:
We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the zeros)x + Product of zeros
Hence, the required polynomial is .
⇒x2−2√x+13=0⇒3x2−32√x+1=0
Page No 66:
Question 19:
We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the zeros)x + Product of zeros
Hence, the required polynomial is .
⇒x2−2√x+13=0⇒3x2−32√x+1=0
Answer:
To find the zeros of the quadratic polynomial we will equate f(x) to 0
Hence, the zeros of the quadratic polynomial f(x) = 6x2 − 3 are .
Page No 66:
Question 20:
To find the zeros of the quadratic polynomial we will equate f(x) to 0
Hence, the zeros of the quadratic polynomial f(x) = 6x2 − 3 are .
Answer:
To find the zeros of the quadratic polynomial we will equate f(x) to 0
Hence, the zeros of the quadratic polynomial are .
Page No 66:
Question 21:
To find the zeros of the quadratic polynomial we will equate f(x) to 0
Hence, the zeros of the quadratic polynomial are .
Answer:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Solving α − β = 1 and α + β = 5, we will get
α = 3 and β = 2
Substituting these values in , we will get
k = 6
Page No 66:
Question 22:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Solving α − β = 1 and α + β = 5, we will get
α = 3 and β = 2
Substituting these values in , we will get
k = 6
Answer:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Page No 69:
Question 1:
By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =
Answer:
(d) is the correct option.
A polynomial in x of degree n is an expression of the form p(x) =ao+a1x+a2x2+...+an xn, where an 0.
Page No 69:
Question 2:
(d) is the correct option.
A polynomial in x of degree n is an expression of the form p(x) =ao+a1x+a2x2+...+an xn, where an 0.
Answer:
It is because in the second term, the degree of
x is −1 and an expression with a negative degree is not a polynomial.
Page No 69:
Question 3:
It is because in the second term, the degree of
x is −1 and an expression with a negative degree is not a polynomial.
Answer:
Page No 69:
Question 4:
Answer:
Page No 69:
Question 5:
Answer:
Page No 69:
Question 6:
Answer:
Page No 70:
Question 7:
Answer:
Page No 70:
Question 8:
Answer:
Page No 70:
Question 9:
Answer:
Page No 70:
Question 10:
Answer:
Multiply by 10, we get
Page No 70:
Question 11:
Multiply by 10, we get
Answer:
Page No 70:
Question 12:
Answer:
Page No 70:
Question 13:
Answer:
Page No 70:
Question 14:
Answer:
Page No 70:
Question 15:
Answer:
Page No 70:
Question 16:
Answer:
Page No 70:
Question 17:
Answer:
Page No 70:
Question 18:
Answer:
Page No 71:
Question 19:
Answer:
Page No 71:
Question 20:
Answer:
Page No 71:
Question 21:
Answer:
Page No 71:
Question 22:
Answer:
Page No 71:
Question 23:
Answer:
Page No 71:
Question 24:
Answer:
Page No 71:
Question 25:
Answer:
Page No 71:
Question 26:
Answer:
Page No 71:
Question 27:
Answer:
Page No 71:
Question 28:
Answer:
Page No 75:
Question 1:
Answer:
(b) 3,-1
Here,
Page No 75:
Question 2:
(b) 3,-1
Here,
Answer:
(a) −1
Here,
Comparing the given polynomial with , we get:
Page No 75:
Question 3:
(a) −1
Here,
Comparing the given polynomial with , we get:
Answer:
(c)
Here,
Comparing the given polynomial with , we get:
It is given that are the roots of the polynomial.Also, =
Page No 75:
Question 4:
(c)
Here,
Comparing the given polynomial with , we get:
It is given that are the roots of the polynomial.Also, =
Answer:
(c) Let the zeroes of the polynomial be .
Here, p
Comparing the given polynomial with , we get:
a = 4, b = −8k and c = 9
Now, sum of the roots
Page No 75:
Question 5:
(c) Let the zeroes of the polynomial be .
Here, p
Comparing the given polynomial with , we get:
a = 4, b = −8k and c = 9
Now, sum of the roots
Answer:
Here, p
Page No 75:
Question 6:
Here, p
Answer:
Page No 75:
Question 7:
Answer:
It is given that the two roots of the polynomial are 2 and −5.
Let
Now, sum of the zeroes, = 2 + (−
5) = −
3Product of the zeroes, = 2−
5 = −
10∴ Required polynomial =
Page No 75:
Question 8:
It is given that the two roots of the polynomial are 2 and −5.
Let
Now, sum of the zeroes, = 2 + (−
5) = −
3Product of the zeroes, = 2−
5 = −
10∴ Required polynomial =
Answer:
The given polynomial and its roots are .
Page No 75:
Question 9:
The given polynomial and its roots are .
Answer:
Let p
Page No 75:
Question 10:
Let p
Answer:
Given:
Sum of the zeroes = −5
Product of the zeroes = 6
∴ Required polynomial =
Page No 75:
Question 11:
Given:
Sum of the zeroes = −5
Product of the zeroes = 6
∴ Required polynomial =
Answer:
Page No 75:
Question 12:
Answer:
Page No 75:
Question 13:
Answer:
Page No 75:
Question 14:
Answer:
Comparing the polynomial with , we get:
Page No 75:
Question 15:
Comparing the polynomial with , we get:
Answer:
Page No 75:
Question 16:
Answer:
Page No 76:
Question 17:
Answer:
Page No 76:
Question 18:
Answer:
Page No 76:
Question 19:
Answer:
Page No 76:
Question 20:
Answer:
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