Find the zeroes of the following polynomials by factorisation method.
Q.1. 4x^{2} – 3x – 1
4x^{2} – 3x – 1
Splitting the middle term, we get,
4x^{2} 4x + 1x  1
Taking the common factors out, we get,
4x(x  1) +1(x  1)
On grouping, we get,
(4x + 1)(x  1)
So, the zeroes are,
4x + 1 = 0⇒ 4x = 1 ⇒x = (1/4)
(x1) = 0 ⇒ x = 1
Therefore, zeroes are (1/4) and 1
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x^{2}
α + β = – b/a
1 – 1/4 = – ( 3)/4 = ¾
Product of the zeroes = constant term ÷ coefficient of x^{2}
α β = c/a
1( 1/4) = – ¼
– 1/4 = – 1/4
Q.2. 3x^{2} + 4x – 4
3x^{2} + 4x – 4
Splitting the middle term, we get,
3x^{2} + 6x – 2x – 4
Taking the common factors out, we get,
3x(x + 2) 2(x + 2)
On grouping, we get,
(x + 2)(3x  2)
So, the zeroes are,
x + 2 =0 ⇒ x = 2
3x  2 = 0 ⇒ 3x = 2 ⇒ x = 2/3
Therefore, zeroes are (2/3) and 2
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x^{2}
α + β = – b/a
– 2 + (2/3) = – (4)/3
= – 4/3 = – 4/3
Product of the zeroes = constant term ÷ coefficient of x^{2}
α β = c/a
Product of the zeroes = ( 2) (2/3) = – 4/3
Q.3. 5t^{2} + 12t + 7
5t^{2} + 12t + 7
Splitting the middle term, we get,
5t^{2} + 5t + 7t + 7
Taking the common factors out, we get,
5t (t + 1) +7(t + 1)
On grouping, we get,
(t + 1)(5t + 7)
So, the zeroes are,
t + 1 = 0 ⇒ y = 1
5t+7=0 ⇒ 5t=7⇒t=7/5
Therefore, zeroes are (7/5) and 1
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x^{2}
α + β = – b/a
( 1) + ( 7/5) = – (12)/5
= – 12/5 = – 12/5
Product of the zeroes = constant term ÷ coefficient of x^{2}
α β = c/a
( 1)( 7/5) = 7/5
7/5 = 7/5
Q.4. t^{3} – 2t^{2} – 15t
t^{3} – 2t^{2} – 15t
Taking t common, we get,
t ( t^{2} 2t 15)
Splitting the middle term of the equation t^{2} 2t 15, we get,
t( t^{2} 5t + 3t 15)
Taking the common factors out, we get,
t (t (t5) +3(t5)
On grouping, we get,
t (t+3)(t5)
So, the zeroes are,
t=0
t+3=0 ⇒ t= 3
t 5=0 ⇒ t=5
Therefore, zeroes are 0, 5 and 3
Verification:
Sum of the zeroes = – (coefficient of x^{2}) ÷ coefficient of x^{3}
α + β + γ = – b/a
(0) + ( 3) + (5) = – ( 2)/1
= 2 = 2
Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x^{3}
αβ + βγ + αγ = c/a
(0)( 3) + ( 3) (5) + (0) (5) = – 15/1
= – 15 = – 15
Product of all the zeroes = – (constant term) ÷ coefficient of x^{3}
αβγ = – d/a
(0)( 3)(5) = 0
Q.5. 2x^{2} +(7/2)x +3/4
2x^{2} +(7/2)x +3/4
The equation can also be written as,
8x^{2}+14x+3
Splitting the middle term, we get,
8x^{2}+12x+2x+3
Taking the common factors out, we get,
4x (2x+3) +1(2x+3)
On grouping, we get,
(4x+1)(2x+3)
So, the zeroes are,
4x+1=0 ⇒ x = 1/4
2x+3=0 ⇒ x = 3/2
Therefore, zeroes are 1/4 and 3/2
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x^{2}
α + β = – b/a
( 3/2) + ( 1/4) = – (7)/4
= – 7/4 = – 7/4
Product of the zeroes = constant term ÷ coefficient of x^{2}
α β = c/a
( 3/2)( 1/4) = (3/4)/2
3/8 = 3/8
Q.6. t^{3}  2t^{2}  15t
Let t³  2t²  15t = 0
Taking out common term,
t(t²  2t  15) = 0
t = 0
On factoring,
t²  2t  15 = t²  5t + 3t  15
= t(t  5) + 3(t  5)
= (t + 3)(t  5)
Now, t + 3 = 0
t = 3
Also, t  5 = 0
t = 5Therefore, the zeros of the polynomial are 0, 3 and 5.
We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then
Sum of the roots is α + β = coefficient of x/coefficient of x² = b/a
Product of the roots is αβ = constant term/coefficient of x² = c/a
From the given polynomial,
coefficient of x = (1+2√2)
Coefficient of x² = 2
Constant term = √2
Sum of the roots:
LHS: α + β
= 1/2 + √2
= (1 + 2√2)/2
RHS: coefficient of x/coefficient of x²
= [(1+2√2)/2]
= (1 + 2√2)/2
LHS = RHS
Product of the roots
LHS: αβ
= (1/2)(√2)
= √2/2
RHS: constant term/coefficient of x²
= √2/2
LHS = RHS
Therefore, the zeros of the polynomial are ½ and √2. The relation between the coefficients is b/a
= (1+2√2)/2 and c/a = √2/2.
Q.7. 2x^{2}  (1 + 2√2)s + √2
Given, the polynomial is 2s²  (1+ 2√2)s + √2.
We have to find the relation between the coefficients and zeros of the polynomial
Let 2s²  (1+ 2√2)s + √2 = 0
On factoring,
2s²  s  2√2s + √2 = 0
2s²  2√2s  s + √2 = 0
2s(s  √2 )  1(s  √2) = 0
(2s  1)(s  √2) = 0
Now, 2s  1 = 0
2s = 1
s = 1/2
Also, s  √2 = 0
s = √2
Therefore, the zeros of the polynomial are 1/2 and √2.
Q.8. v^{2} + 4√3v  15
Given, the polynomial is v² + 4√3v  15.
We have to find the relation between the coefficients and zeros of the polynomial
Let v² + 4√3v  15 = 0
On factoring,
v² + 5√3v  √3v  15 = 0
v(v + 5√3)  √3(v + 5√3) = 0
(v  √3)(v + 5√3) = 0
Now, v  √3 = 0
v = √3
Also, v + 5√3 = 0
v = 5√3
Therefore, the zeros of the polynomial are 5√3 and √3.
We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then
Sum of the roots is α + β = coefficient of x/coefficient of x² = b/a
Product of the roots is αβ = constant term/coefficient of x² = c/a
From the given polynomial,
coefficient of x = 4√3
Coefficient of x² = 1
Constant term = 15
Sum of the roots:
LHS: α + β
= √3  5√3
= 4√3
RHS: coefficient of x/coefficient of x²
= 4√3/1
= 4√3
LHS = RHS
Product of the roots
LHS: αβ
= (5√3)(√3)
= 15
RHS: constant term/coefficient of x²
= 15/1
= 15
LHS = RHS
Q.9.
Given, the polynomial is y² + (3√5/2)y  5.
We have to find the relation between the coefficients and zeros of the polynomial
The polynomial can be rewritten as (1/2)[2y² + 3√5y  10].
Let 2y² + 3√5y  10 = 0
On factoring,
2y² + 4√5y  √5y  10 = 0
2y(y + 2√5)  √5(y  2√5) = 0
(y + 2√5)(2y  √5) = 0
Now, y + 2√5 = 0
y = 2√5
Also, 2y  √5 = 0
2y = √5
x = √5/2
Therefore, the zeros of the polynomial are √5/2 and 2√5.
We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then
Sum of the roots is α + β = coefficient of x/coefficient of x² = b/a
Product of the roots is αβ = constant term/coefficient of x² = c/a
From the given polynomial,
coefficient of x = 3√5
Coefficient of x² = 2
Constant term = 10
Sum of the roots:
LHS: α + β
= √5/2  2√5
= (√5  4√5)/2
= 3√5/2
RHS: coefficient of x/coefficient of x²
= 3√5/2
LHS = RHS
Product of the roots
LHS: αβ
= (√5/2)(2√5)
= 10/2
= 5
RHS: constant term/coefficient of x²
= 10/2
= 5
LHS = RHS
Q.10. 7y^{2}11/3y  2/3
Given, the polynomial is 7y²  (11/3)y  (2/3).
We have to find the relation between the coefficients and zeros of the polynomial
The polynomial can be rewritten as (1/3)[21y²  11y  2]
Let (1/3)[21y²  11y  2] = 0
21y²  11y  2 = 0
On factoring,
21y²  14y + 3y  2 = 0
7y(3y  2) + (3y  2) = 0
(7y + 1)(3y  2) = 0
Now, 7y + 1 = 0
7y = 1
y = 1/7
Also, 3y  2 = 0
3y = 2
y = 2/3
Therefore, the zeros of the polynomial are 2/3 and 1/7.
We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then
Sum of the roots is α + β = coefficient of x/coefficient of x² = b/a
Product of the roots is αβ = constant term/coefficient of x² = c/a
From the given polynomial,
coefficient of x = 11
Coefficient of x² = 21
Constant term = 2
Sum of the roots:
LHS: α + β
= 1/7 + 2/3
= (3+14)/21
= 11/21
RHS: coefficient of x/coefficient of x²
= (11)/21
= 11/21
LHS = RHS
Product of the roots
LHS: αβ
= (1/7)(2/3)
= 2/21
RHS: constant term/coefficient of x²
= 2/21
LHS = RHS
Q.1. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
(i) (–8/3), 4/3
(ii) 21/8, 5/16
(iii) 2√3, 9
(iv) (3/(2√5)), ½
(i) Sum of the zeroes = – 8/3
Product of the zeroes = 4/3
P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)
Then, P(x)= x^{2} – (8x)/3 + 4/3
P(x)= 3x^{2} + 8x + 4
Using splitting the middle term method,
3x^{2} + 8x + 4 = 0
3x^{2} + (6x + 2x) + 4 = 0
3x^{2} + 6x + 2x + 4 = 0
3x(x + 2) + 2(x + 2) = 0
(x + 2)(3x + 2) = 0
⇒ x = 2, 2/3
(ii) Sum of the zeroes = 21/8
Product of the zeroes = 5/16
P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)
Then, P(x)= x^{2} – 21x/8 + 5/16
P(x)= 16x^{2} – 42x + 5
Using splitting the middle term method,
16x^{2} – 42x + 5 = 0
16x^{2} – (2x + 40x) + 5 = 0
16x^{2} – 2x – 40x + 5 = 0
2x (8x – 1) – 5(8x – 1) = 0
(8x – 1)(2x – 5) = 0
⇒ x = 1/8, 5/2
(iii) Sum of the zeroes = – 2√3
Product of the zeroes = – 9
P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)
Then, P(x) = x^{2} – (2√3x) – 9
Using splitting the middle term method,
x^{2} + 2√3x – 9 = 0
x^{2} + (3√3x – √3x) – 9 = 0
x(x + 3√3) – √3(x + 3√3) = 0
(x – √3)(x + 3√3) = 0
⇒ x = √3, 3√3
(iv) Sum of the zeroes = 3/2√5x
Product of the zeroes = – ½
P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)
Then, P(x)= x^{2} (3/2√5x) – ½
P(x)= 2√5x^{2} + 3x – √5
Using splitting the middle term method,
2√5x^{2} + 3x – √5 = 0
2√5x^{2} + (5x – 2x) – √5 = 0
2√5x^{2} – 5x + 2x – √5 = 0
√5x (2x + √5) – (2x + √5) = 0
(2x + √5)(√5x – 1) = 0
⇒ x = 1/√5, √5/2
Q.2. Given that the zeroes of the cubic polynomial x^{3} – 6x^{2} + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.
Given that a, a + b, a + 2b are roots of given polynomial x³  6x² + 3x + 10
Sum of the roots ⇒ a + 2b + a + a + b = coefficient of x²/ coefficient of x³
⇒ 3a+3b = (6)/1 = 6
⇒ 3(a+b) = 6
⇒ a+b = 2 ........(1)b = 2a
Product of roots ⇒ (a+2b)(a+b)a = constant/coefficient of x³
⇒ (a+b+b)(a+b)a = 10/1
Substituting the value of a+b=2 in it
⇒ (2+b)(2)a = 10
⇒ (2+b)2a = 10
⇒ (2+2a)2a = 10
⇒ (4a)2a = 10
⇒ 4aa² = 5
⇒ a²4a5 = 0
⇒ a²5a+a5 = 0
⇒ (a5)(a+1) = 0
a5 = 0 or a+1 = 0
a = 5 a = 1
a = 5, 1 in (1) a+b = 2
When a = 5, 5+b=2 ⇒ b=3
a = 1, 1+b=2 ⇒ b= 3
∴ If a=5 then b= 3
or
If a= 1 then b=3
Q.3. Given that √2 is a zero of the cubic polynomial 6x^{3} + √2 x^{2} – 10x – 4√2 , find its other two zeroes.
Given, √2 is one of the zero of the cubic polynomial.
Then, (x√2) is one of the factor of the given polynomial p(x) = 6x³+√2x²10x 4√2.
So, by dividing p(x) by x√2
6x³+√2x²10x4√2= (x√2) (6x² +7√2x + 4)
By splitting the middle term,
We get,
(x√2) (6x² + 4√2x + 3√2x + 4)
= (x√2) [2x(3x+2√2) + √2(3x+2√2)]
= (x√2) (2x+√2) (3x+2√2)
To get the zeroes of p(x),
Substitute p(x)= 0
(x√2) (2x+√2) (3x+2√2)= 0
x= √2, x= √2/2, x= 2√2/3
Hence, the other two zeroes of p(x) are √2/2 and 2√2/3
Q.4. Find k so that x^{2} + 2x + k is a factor of 2x^{4} + x^{3} – 14x^{2 }+ 5x + 6. Also find all the zeroes of the two polynomials.
Given, p(x) = 2x⁴ + x³  14 x² + 5x + 6.
g(x) = x² + 2x + k
We have to find the zeros of the polynomial.
The division algorithm states that given any polynomial p(x) and any nonzero
polynomial g(x), there are polynomials q(x) and r(x) such that
p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).
Let r(x) = 0
So, p(x) = g(x) q(x)
By using long division,
So, q(x) = 2x²  3x  8  2k
r(x) = (21+7k)x + (2k² + 8k + 6)
By comparing the coefficients of (21+7k)x and 2k² + 8k + 6
2k² + 8k + 6 = 0
2(k² + 4k + 3) = 0
k² + 3k + k + 3 = 0
k(k + 3) + (k + 3) = 0
(k + 1)(k + 3) = 0
Now, k + 1 = 0
k = 1
Also, k + 3 = 0
k = 3
So, k = 1, 3.
When k = 1,
21 + 7k = 0
= 21 + 7(1)
= 21  7
= 14
21 + 7k is not equal to zero.
So, k = 1 is neglected.
When k = 3,
21 + 7k = 0
= 21 + 7(3)
= 21  21
= 0
Therefore, the value of k is 3.
Now, g(x) = x² + 2x  3
x² + 2x  3 = 0
x²  x + 3x  3 = 0
x(x  1) + 3(x  1) = 0
(x + 3)(x  1) = 0
Now, x + 3 = 0
x = 3
Also, x  1 = 0
x = 1
Now, q(x) = 2x²  3x  8  2k
= 2x²  3x  8  2(3)
= 2x²  3x  8 + 6
= 2x²  3x  2
On factoring,
2x²  3x  2 = 0
2x²  4x + x  2 = 0
2x(x  2) + (x  2) = 0
(2x + 1)(x  2) = 0
Now, 2x + 1 = 0
2x = 1
x = 1/2
Also, x  2 = 0
x = 2
We know that g(x) and q(x) are the factors of p(x).
So, the zeros of g(x) and q(x) will be the zeros of p(x).
Therefore, the zeros of p(x) = 3, 1/2, 1 and 2.
Q.5. Given that x – √5 is a factor of the cubic polynomial x^{3} – 3√5x^{2} + 13x – 3√5, find all the zeroes of the polynomial.
Give, the cubic polynomial p(x) = x³  3√5x² + 13x  3√5
g(x) = x  √5
We have to find all the zeros of the polynomial.
By using long division,
The quotient is g(x) = x²  2√5x + 3
On factoring,
x²  2√5x + 3 = 0
Using the quadratic formula,
x= [b ± √b²  4ac]/2a
Here, a = 1, b = 2√5 and c = 3
x = [2√5 ± √(2√5)²  4 (1) (3)]/ 2(1)
x = [2√5 ± √20  12]/ 2
x = [2√5 ± √8]/ 2
x = [2√5 ± 2√2]/ 2
Taking out common terms,
x = 2[√5 ± √2]/ 2
x = √5 ± √2
Therefore, the zeros are √5, √5 ± √2.
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