factorise (2a+3)³ + (3a-2³) + (-1-5a³)
Answers
Answer:
Question
Question
Q. Given that the zeros of the cubic polynomial x³-6x²+3x+10 are the form of a, a+b and a+2b for some real number a and b find the values of a and b as well as the zeros of the given polynomial.
Given :-
a, a+b and a+2b are roots of x³-6x²+3x+10
Required to find :-
value of a and b and zeros of polynomial
\huge \huge \bf {\: \pmb{Solution}}
Solution
Solution
given that a, a+b and a+2b are roots of x³-6x²+3x+10
we know that sum of roots = \frac{-cofficient~of~x²}{coffiecient ~of~x³ }
coffiecient of x³
−cofficient of x²
= a + 2b + a + a + b = \frac{-(-6)}{1}
1
−(−6)
= 3a + 3b = 6
= 3 ( a + b ) = 6
= a + b = \frac{6}{3}
3
6
= a + b = 2 --- ( 1 )
= b = 2 - a
we know that product of roots = \frac{-constant}{coffiecient ~of~x³ }
coffiecient of x³
−constant
( a + 2b )( a + b )a = \frac{-10}{1}
1
−10
a + b + b ) ( a + b ) 2 = \frac{-10}{1}
1
−10
substituting equation 1 here,
= ( 2 + b ) (2)a = -10
= ( 2 + b ) 2a = -10
= ( 4 - a )2a = -10
= 4a - a² = \frac{-10}{5}
5
−10
= a² - 4a + 5 = 0
by splitting the middle term
= a² - 5a + a - 5 = 0
= a ( a - 5 ) + 1 ( a - 5 ) = 0
= ( a + 1 ) ( a - 5 ) = 0
→ a + 1 = 0
• a = -1
→ a - 5 = 0
• a = 5
→ when a = -1
= a + b = 2
= -1 + b = 2
= b = 2 + 1
= b = 3
• a = -1 , b = 3
→ when a = 5
= a + b = 2
= 5 + b = 2
= b = 2 - 5
= b = -3
• a = 5 , b = -3
Now, zeros of the given polynomial.
→ a = -1 & b = 3
• a = -1
• a+b = (-1) + 3 = 2
• a+2b = (-1) + 2 ( 3 ) = -1 + 6 = 5
→ a = 5 & b = -3
• a = 5
• a + b = 5 + (- 3 )= 2
• a + 2b = 5 + 2(-3) = 5 + (-6 )= -1