Polynomials Ex 2.4 Solutions
Question - 1 : - For each of thefollowing, find a quadratic polynomial whose sum and product respectively ofthe zeroes are as given. Also find the zeroes of these polynomials byfactorisation.
(i) (–8/3), 4/3
(ii) 21/8, 5/16
(iii) -2√3, -9
(iv) (-3/(2√5)), -½
Answer - 1 : -
Solution:
(i) Sum of the zeroes= – 8/3
Product of the zeroes= 4/3
P(x) = x2 – (sum of the zeroes) + (product of the zeroes)
Then, P(x)= x2 – (-8x)/3 + 4/3
P(x)= 3x2 + 8x + 4
Using splitting themiddle term method,
3x2 + 8x + 4 = 0
3x2 + (6x + 2x) + 4 = 0
3x2 + 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) = x2 – (sum of the zeroes) + (product of the zeroes)
Then, P(x)= x2 – 21x/8 + 5/16
P(x)= 16x2 – 42x + 5
Using splitting themiddle term method,
16x2 – 42x + 5 = 0
16x2 – (2x + 40x) + 5 = 0
16x2 – 2x – 40x + 5 = 0
2x (8x – 1) – 5(8x –1) = 0
(8x – 1)(2x – 5) = 0
⇒ x = 1/8, 5/2
(iii) Sum of thezeroes = – 2√3
Product of the zeroes= – 9
P(x) = x2 – (sum of the zeroes) + (product of the zeroes)
Then, P(x) = x2 – (-2√3x) – 9
Using splitting themiddle term method,
x2 + 2√3x – 9 = 0
x2 + (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) = x2 – (sum of the zeroes) + (product of the zeroes)
Then, P(x)= x2 -(-3/2√5x) – ½
P(x)= 2√5x2 + 3x – √5
Using splitting themiddle term method,
2√5x2 + 3x – √5 = 0
2√5x2 + (5x – 2x) – √5 = 0
2√5x2 – 5x + 2x – √5 = 0
√5x (2x + √5) – (2x +√5) = 0
(2x + √5)(√5x – 1) = 0
⇒ x = 1/√5, -√5/2
Question - 2 : - Given that the zeroesof the cubic polynomial x3 – 6x2 + 3x + 10 are of the form a, a + b, a +2b for some real numbers a and b, findthe values of a and b as well as the zeroesof the given polynomial.
Answer - 2 : -
Solution:
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/ coefficientof x³
⇒ 3a+3b = -(-6)/1 = 6
⇒ 3(a+b) = 6
⇒ a+b = 2 ——— (1) b =2-a
Product of roots ⇒ (a+2b)(a+b)a = -constant/coefficient of x³
⇒ (a+b+b)(a+b)a = -10/1
Substituting the valueof a+b=2 in it
⇒ (2+b)(2)a = -10
⇒ (2+b)2a = -10
⇒ (2+2-a)2a = -10
⇒ (4-a)2a = -10
⇒ 4a-a² = -5
⇒ a²-4a-5 = 0
⇒ a²-5a+a-5 = 0
⇒ (a-5)(a+1) = 0
a-5 = 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
Question - 3 : - Given that √2 is a zero of the cubic polynomial 6x3 + √2 x2 – 10x – 4√2 , find its other two zeroes.
Answer - 3 : -
Given, √2 is one ofthe zero of the cubic polynomial.
Then, (x-√2) is one ofthe factor of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.
So, by dividing p(x)by x-√2
6x³+√2x²-10x-4√2=(x-√2) (6x² +7√2x + 4)
By splitting themiddle 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 ofp(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 twozeroes of p(x) are -√2/2 and -2√2/3
Question - 4 : - Verify that the numbers given alongside of thecubic polynomials below are their zeroes. Also verify the relationship betweenthe zeroes and the coefficients in each case:
(i) 2x3+x2-5x+2; -1/2,1, -2 (ii) x3-4x2+5x-2 ;2, 1, 1
Answer - 4 : -
(i) 2x3+x2-5x+2; -1/2,1, -2
Solution:
Given, p(x) = 2x3+x2-5x+2
And zeroes for p(x) are = 1/2, 1, -2
∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2
= (1/4)+(1/4)-(5/2)+2
= 0
p(1)= 2(1)3+(1)2-5(1)+2
= 0
p(-2)= 2(-2)3+(-2)2-5(-2)+2
= 0
Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.
Now, comparing the given polynomial withgeneral expression, we get;
∴ ax3+bx2+cx+d
= 2x3+x2-5x+2
a=2, b=1, c= -5 and d = 2
As we know, if α, β, γ are the zeroes of thecubic polynomial ax3+bx2+cx+d , then;
α +β+γ = –b/a
αβ+βγ+γα = c/a
α βγ = – d/a.
Therefore, putting the values of zeroes of thepolynomial,
α+β+γ = ½+1+(-2)
= -1/2
= –b/a
αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2)
= -5/2
= c/a
α β γ = ½×1×(-2)
= -2/2
= -d/a
Hence, the relationship between the zeroes andthe coefficients are satisfied.
(ii) x3-4x2+5x-2 ;2, 1, 1
Solution:
Given, p(x) = x3-4x2+5x-2
And zeroes for p(x) are 2,1,1.
∴ p(2)= 23-4(2)2+5(2)-2
= 0
p(1) = 13-(4×12 )+(5×1)-2
= 0
Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2
Now, comparing the given polynomial withgeneral expression, we get;
∴ ax3+bx2+cx+d = x3-4x2+5x-2
a = 1, b = -4, c = 5 and d = -2
As we know, if α, β, γ are the zeroes of thecubic polynomial ax3+bx2+cx+d , then;
α + β + γ = –b/a
αβ + βγ + γα = c/a
α β γ = – d/a.
Therefore, putting the values of zeroes of thepolynomial,
α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a
αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a
αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a
Hence, the relationship between the zeroes andthe coefficients are satisfied.
Question - 5 : - Find a cubic polynomial with the sum, sum ofthe product of its zeroes taken two at a time, and the product of its zeroes as2, –7, –14 respectively.
Answer - 5 : -
Solution:
Let us consider the cubic polynomial is ax3+bx2+cx+dand the values of the zeroes of the polynomials be α, β, γ.
As per the given question,
α+β+γ = -b/a
= 2/1
αβ +βγ+γα = c/a
= -7/1
α βγ = -d/a
= -14/1
Thus, from above three expressions we get thevalues of coefficient of polynomial.
a = 1, b = -2, c = -7, d = 14
Hence, the cubic polynomial is x3-2x2-7x+14
Question - 6 : - If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.
Answer - 6 : -
Solution:
We are given with the polynomial here,
p(x) = x3-3x2+x+1
And zeroes are given as a – b, a, a + b
Now, comparing the given polynomial withgeneral expression, we get;
∴px3+qx2+rx+s = x3-3x2+x+1
p = 1, q = -3, r = 1 and s = 1
Sum of zeroes = a – b + a + a + b
-q/p = 3a
Putting the values q and p.
-(-3)/1 = 3a
a=1
Thus, the zeroes are 1-b, 1, 1+b.
Now, product of zeroes = 1(1-b)(1+b)
-s/p = 1-b2
-1/1 = 1-b2
b2 = 1+1 = 2
b = √2
Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1.
Question - 7 : - two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, findother zeroes.
Answer - 7 : - 
Question - 8 : - _If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x +a, find k and a.
Answer - 8 : -
