answer it plz plz plz
Answers
10) Given
A polynomial ax³+3x²-bx-6
And their zeroes is -6 and -1
To find
value of a,b and third zeroes
Solution
Let the third zeroes of the polynomial be
putting the value of zeroes X=(-1,-2) in polynomial
First X= -1
f(X}=ax³+3x²-bx-6 =0
f(-1)=-a+3+b-6
=>a+b=3______(1)
second X = -2
F(X)=ax³+3x²-bx-6=0
f(-6)=-8a+12+2b-6
=>-8a+2b=-6
=>4a-b=3
From equation (1) and (2)
a+b=3
4a-b=3
___________
5a=6
a=6/5
b=9/5
Now we know
\alpha.\beta\gamma=\frac{-d}{a}
2\gamma=5
8) Since √3 and -√3 are zeros , x-√3 and x+√3 are factors of the polynomial x^4-3x³-x²+9x-6 .
(x+√3)(x-√3) = x²-(√3)² [ (a+b)(a-b)=a²-b²]
= x²- 3
Now divide the polynomial x^4-3x³-x²+9x-6 by (x²-3)
=>
x² - 3x + 2
-----------------------
x²-3 ║ x^4-3x³-x²+9x-6
x^4 -3x²
- +
--------------------------
0-3x³+2x²+9x-6
-3x³ +9x
+ -
---------------------------
0+2x²+0-6
2x² -6
- +
---------------------------
0 0
=>x^4-3x³-x²+9x-6 = (x²-3x+2)(x²- 3)
=>(x²-3x+2) = x²-2x-x+2)
=x(x-2)-1(x-2)
=(x-2)(x-1)
=>x=2,1
=>x^4-3x³-x²+9x-6 = (x-2)(x-1)(x+√3)(x-√3)
Therefore the zeros are 2,1,√3 and -√3
14) ( see question 14 I have copied from internet it related to your question pls change no 15 instead of 6)
14) The relation between the zeroes and coefficients of a quadratic equation are as follows:
For every polynomial of form ax²+bx+c
Now the given polynomial is
P(x) =x²+8x+6
So for this polynomial
Now,
We are asked a polynomial whose zeroes are
Now let the polynomial be f(x)
So,
For f(x) the sum of the zeroes is
Now, for the product of zeroes of f(x)
Now,
For f(x)
Sum of roots = 44/6
Product of zeroes = - 1/6
The structure of the polynomial is
K(x²-(sum of zeroes) x +(product of zeroes)
So, f(x) = k(x²-44/6-1/6)
Or, f(x) = 6x²-44x-1
When k = 6
15) I don't know sorry sis
pls follow me and mark me brainlest
