find the zeros of the polynomial p(x)=x³-12x²+39x-28, if it is given that the zeroes are a-d, a and a+d
Answers
Answer :
The zeroes are 1,4 and 7.
Given :
→ p(x) = x³-12x²+39x-28
→ the zeroes are a-d, a and a+d
To find :
→ Zeroes of the polynomial
Solution :
We must be knowing, the relationship between zeroes and coefficients to find the zeroes.
For a polynomial of the form ax³+bx²+cx+d
The relationship between zeroes and coefficients of a cubic polynomial :
⇒ Sum of zeroes = -b/a
⇒ Product of zeroes = -d/a
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Given polynomial,
p(x) = x³-12x²+39x-28
║a = 1, b = -12, c = 39, d = -28
>> Sum of zeroes = -(-12)/1
a-d + a + a+d = 12
3a = 12
a = 12/3
a = 4
>> Product of zeroes = -(-28)/1
(a-d)(a)(a+d) = 28
(a² - d²)(a) = 28
(4² - d²)(4) = 28
16 - d² = 28/4
16 - d² = 7
d² = 16-7
d² = 9
d = √9
d = 3
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>>> The zeroes are a-d , a , a+d
⇒ a-d = 4-3 = 1
⇒ a = 4
⇒ a+d = 4+3 = 7
The zeroes of the polynomial p(x) = x³-12x²+39x-28 are 1,4 and 7