The Polynomial p(x)=x^4-2x^3+3x²-ax+3a-7when divided by x+1 leaves remainder 19.Find value of a and find the remainder when p(x) is divided by x+2 14. If both X-2 and x-1/2 are factors of px²+5x+r,show that p=r.
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Answers
Step-by-step explanation:
P(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7. Thus, P(-1) = (-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + 3a - 7. ∴ Value of a is 5.
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Answer:
p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7.
Divisor = x + 1
x + 1 = 0
x = -1
On substituting the value of x = – 1 in p(x), we obtain,
p(-1) = (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + 3a – 7.
19 = 1 + 2 + 3 + a + 3a – 7
19 = 6 – 7 + 4a
4a – 1 = 19
4a = 20
a = 5
Since, a = 5.
The polynomial can be expressed as
p(x) = x⁴ – 2x³ + 3x² – (5)x + 3(5) – 7
p(x) = x⁴ – 2x³ + 3x² – 5x + 15 – 7
p(x) = x⁴ – 2x³ + 3x² – 5x + 8
Also, according to the given details
When the polynomial obtained is divided by (x + 2),
So
x + 2 = 0
x = – 2
On substituting the value of x = – 2 in p(x), we get,
p(-2) = (-2)⁴ – 2(-2)³ + 3(-2)² – 5(-2) + 8
⇒ p(-2) = 16 + 16 + 12 + 10 + 8
⇒ p(-2) = 62
Therefore, the remainder = 62.
Answer
The value of a=5
The remainder when p(x) is divided by x + 2=62
Step-by-step explanation:
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