find the factors of polynomial 6x^2 - 3
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Step-by-step explanation:
Given
f(x) = 6x² – 7x – 3
To find the zeros
Let us put f(x) = 0
⇒ 6x² – 7x – 3 = 0
⇒ 6x² – 9x + 2x – 3 = 0
⇒ 3x(2x – 3) + 1(2x – 3) = 0
⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0
x = 3/2
⇒ 3x + 1 = 0
⇒ x = -1/3
It gives us 2 zeros, for x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, for verification
Sum of zeros = – coefficient of x / coefficient of x2
3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.
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