find zeros of the polynomials x(x-28)-15
Answers
Answer:
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Step-by-step explanation:
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Answer:
f(x) = x² - 3x - 28
To find zero
put f(x) = 0
x² - 3x - 28 = 0
x² - 7x + 4x - 28 = 0
x ( x - 7 ) + 4 (x - 7 ) = 0
( x - 7 )( x + 4 ) = 0
x = 7 & -4
let α = 7 and β = -4
Now to verify the relation
sum of zeroes = α + β = 7 + (-4) = 3
\frac{-coefficient\:of\:x}{Coefficient\:of\:x^2}=\frac{-(-3)}{1}=3
Coefficientofx
2
−coefficientofx
=
1
−(−3)
=3
Product of zeroes = α . β = 7 × (-4) = -28
\frac{constant\:term}{Coefficient\:of\:x^2}=\frac{-28}{1}=-28
Coefficientofx
2
constantterm
=
1
−28
=−28
sum of zeroes = \frac{-coefficient\:of\:x}{Coefficient\:of\:x^2}
Coefficientofx
2
−coefficientofx
Product of zeroes = \frac{constant\:term}{Coefficient\:of\:x^2}
Coefficientofx
2
constantterm
Hence proved