Find the zero of the polynomial and verify the relationship between the zeros and their coefficients: 3√2 x∧2 + 13x +6√2
Answers
Answer:
√2 x² + 13 x + 6√2
= 3√2 x² + 9x + 4x + 6√2
= 3x(√2 x + 3) + 2√2(√2x + 3)
= (√2x + 3)(3x + 2√2)
Zeroes of the polynomial:
(3x + 2√2) = 0
⇒ x = 2√2/3
(√2x + 3) = 0
⇒ x = -3/√2
Sum of the roots = 2√2/3 -3/√2 = -13/3√2
Product of the roots = 2√2/3 × 3/√2 = 2
Sum of the roots = -b/a = -13/3√2√2 x² + 13 x + 6√2
= 3√2 x² + 9x + 4x + 6√2
= 3x(√2 x + 3) + 2√2(√2x + 3)
= (√2x + 3)(3x + 2√2)
Zeroes of the polynomial:
(3x + 2√2) = 0
⇒ x = 2√2/3
(√2x + 3) = 0
⇒ x = -3/√2
Sum of the roots = 2√2/3 -3/√2 = -13/3√2
Product of the roots = 2√2/3 × 3/√2 = 2
Sum of the roots = -b/a = -13/3√2
Product of the roots = c/a = 6√2 / 3√2 = 2.
4.4
Product of the roots = c/a = 6
(This photo represents the relationship between zeroes and their coefficient)
3√2x²+13x+6√2
= 3√2x²+9x+4x+6√2. ( By middle term splitting)
= 3x(√2x+3)+ 2√2(√2x+3)
= (3x+2√2)(√2x+3)
Zeroes of the polynomial are: -2√2/3 and -3/√2
