1. Find the zeroes of the following quadratic polynomials and verify the relationships between the zeroes
and the coefficients of the polynomials:- (a): p(x) = 8x² – 19x – 15; (b): q(x) = 4 13 x² + 5x – 2 13; (c): f
(x) = 5x - 473 +2 V3 x?.
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes
respectively:-
(i) 2), -4 (ii)
一、
-YE'Yuva
(iii) V5,-2 (
iv) - VR. 2
3. Verify that 3, -1 and -, are zeroes of the polynomial p(x) = 3x?- 5x² – 11x – 3. Then, verify the
relationships between the zeroes and its coefficients.
4. For what value of K is 4 a zero of f(x) = x² + kx + 4?
5. Find the quotient and remainder when p(x) is divided by f(x).
(i) p(x) = 6x' + 11x² – 39x – 65, 9(x) = x² – 1 + x
p(x) = 4 + 9x² - 4x², 9(x) = x + 3x² - 1
(111) p(x) = 30x* – 82x² + 11x + 48 – 12x, q(x) = 3x² + 2x - 4
6. What must be subtracted from 8x4 + 14xy – 2x² + 7x – 8 so that the resulting polynomial is exactly
divisible by 4x² + 3x – 2?
7. What must be added to 4x4 + 2x² – 2x² + x - 1, so that the resulting polynomial is divisible by x² + 2x –
3?
8. If -2 is a zero of f(x) = xº + 13x² + 32x + 20, find its other zeroes.
9. v3 and - 13 are zeroes of f(x) = x* – 3x’ – x² + 9x – 6. Find all the zeroes of p(x).
10. Obtain all zeroes of the polynomial p(x) = 2x4 + x’ – 14x² – 19x – 6, if two of its zeroes are -1 and -2.
11. Find all the zeroes of f(x) = 2x4 - 2x - 7x² + 3x + 6 if two of its zeroes are 13 and - 15/.
12. Find all values of p and q so that 1, -2 are zeroes of the polynomial f(x)= x' + 10x² + px + q.
13. If p(x)=2x* + 3x - 3x² – 2x + 5 is divided by 2x² + 3x – 1,then the remainder is x - a. Find a.
14. On dividing f(x) = 2x’ – 5x² + 4x – 8 by g(x), the quotient and the remainder are (2x – 9) and 24x - 17,
respectively. Find g(x).
V3,
Answers
Answer:
Given polynomial is 5x
2
−8x−4
Here, a=5,b=−8 and c=−4
5x
2
−8x−4=5x
2
−10x+2x−4
=5x(x−2)+2(x−2)
=(x−2)(5x+2)
So, the value of 5x
2
−8x−4 is zero when x=2 or x=
5
−2
Threrfore , the zeroes of 5x
2
−8−4 are 2 and
5
−2
.
Now, sum of zeroes =2+(
5
−2
)=
5
8
=
a
−b
product of zeroes =2×(
5
−2
)=
5
−4
=
a
c
Hence verified.
Answer:
Given expresion is f(x) = 6x^2 − 3 − 7x
=> f(x) = 6x²-7x-3
=> f(x) = 6x²+2x-9x-3
=> f(x) = 2x(3x+1)-3(3x+1)
=> f(x) = (3x+1)(2x-3)
Factorization of f(x) = (3x+1)(2x-3)
To get zeroes we write f(x) = 0
=> (3x+1)(2x-3)=0
=> 3x+1 = 0 or 2x-3 = 0
=> 3x = -1 or 2x = 3
=> x = -1/3 or x = 3/2
Zeroes are -1/3 and 3/2
e)
Given expresion is p(x) = x^2 + 2√2x − 6
=> p(x) = x² + 2√2x − 6
=> p(x) = x²+3√2x-√2x - 6
=> p(x) = x(x+3√2)-√2(x+3√2)
=> p(x) = (x+3√2)(x-√2)
Factorization of p(x) = (x+3√2)(x-√2)
To get zeroes we write p(x) = 0
=> (x+3√2)(x-√2)=0
=> x+3√2 = 0 or x-√2= 0
=> x = -3√2 or x = √2
Zeroes are -3√2 and √2
f)
Given expresion is q(x) = √3x^2 + 10x + 7√3
=> q(x) =√3x²+10x+7√3
=> q(x) =√3x²+3x+7x+7√3
=> q(x) =√3x(x+√3)+7(x+√3)
=> q(x) = (x+√3)(√3x+7)
Factorization of q(x) = (x+√3)(√3x+7)
To get zeroes we write q(x) = 0
=> (x+√3)(√3x+7) = 0
=> x+√3 = 0 or √3x+7 = 0
=> x = -√3 or √3x = -7
=> x = -√3 or x = -7/√3
Zeroes are -√3 and -7/√3
g)
Given expresion is f(x) = x^2 − (√3 + 1)x + √3
=> f(x) = x²-√3x-x+√3
=> f(x) = x²-x-√3x+√3
=> f(x) = x(x-1)-√3(x-1)
=> f(x) = (x-1)(x-√3)
Factorization of f(x) = (x-1)(x-√3)
To get zeroes we write f(x) = 0
=> (x-1)(x-√3) = 0
=> x-1 = 0 or x-√3= 0
=> x = 1 or x = √3
Zeroes are 1 and √3
h)
Given expresion is g(x) = a(x^2 + 1) − x(a^2 + 1
=> g(x) = a(x² + 1) − x(a² + 1)
=> g(x) = ax²+a-a²x-x
=> g(x) = ax²-a²x-x+a
=> g(x)= ax(x-a)-1(x-a)
=> g(x) = (x-a)(ax-1)
Factorization of g(x) = (x-a)(ax-1)
To get zeroes we write g(x) = 0
=> (x-a)(ax-1) = 0
=>x-a = 0 or ax-1 = 0
=> x = a or ax = 1
=> x = 1 or x = 1/a
Zeroes are 1 and 1/a
Step-by-step explanation: