solve this please.. send a photo
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akay75:
hi
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put x= -1/2 in p(x)
p(-1/2)=(-1/2)^4+3(-1/2)^3-2(-1/2)^2+5(-1/2)+3
p(-1/2)=1/16+-3/8-2/4-5/2+3
p(-1/2)=1/16-6/16-8/16-40/16+48/16
p(-1/2)=-7/16t
therefore x+1/2is not a doctor of p(x)
p(-1/2)=(-1/2)^4+3(-1/2)^3-2(-1/2)^2+5(-1/2)+3
p(-1/2)=1/16+-3/8-2/4-5/2+3
p(-1/2)=1/16-6/16-8/16-40/16+48/16
p(-1/2)=-7/16t
therefore x+1/2is not a doctor of p(x)
Answered by
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hey mate!
let, x+1/2 = 0 for finding that it is the 0 of the polynomial or not!
then, x = -1/2
p(x) = x⁴ + 3x³ - 2x²+ 5x + 3
so, p (-1/2) = (-1/2)⁴ + 3(-1/2)³ - 2(-1/2)² + 5(-1/2) + 3
=> p(-1/2) = (1/16) + 3 * (-1/8) - 2*1/4 - 5/2 + 3
=> p(-1/2) = (1/16) -3/8 -1/2 -5/2 +3/1
=> p(-1/2) = (1-6-8-40+48)/16 { by taking LCM as 16}
=> p(-1/2)= -5/16 ≠ 0
thus, (x + 1/2) is not a factor of p(x) = x⁴ + 3x³ - 2x²+ 5x + 3.
let, x+1/2 = 0 for finding that it is the 0 of the polynomial or not!
then, x = -1/2
p(x) = x⁴ + 3x³ - 2x²+ 5x + 3
so, p (-1/2) = (-1/2)⁴ + 3(-1/2)³ - 2(-1/2)² + 5(-1/2) + 3
=> p(-1/2) = (1/16) + 3 * (-1/8) - 2*1/4 - 5/2 + 3
=> p(-1/2) = (1/16) -3/8 -1/2 -5/2 +3/1
=> p(-1/2) = (1-6-8-40+48)/16 { by taking LCM as 16}
=> p(-1/2)= -5/16 ≠ 0
thus, (x + 1/2) is not a factor of p(x) = x⁴ + 3x³ - 2x²+ 5x + 3.
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