x(cube)-7x(square)+6x+4 is divided by (x-6)
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x^3 - 7x^2 + 6x + 4 ÷ (x - 6)
According to Remainder theorem,
If a polynomial p(x) divided by x-a then remainder is equal to p(a)..
Let x^3 - 7x^2 + 6x + 4 as p(x) .
Hence value of a will be 6.
Hence remainder = p(6)
=) p(6) = x^3 - 7x^2 + 6x + 4
= 6^3 - 7* 6^2 + 6*6 + 4
= 216 - 252 + 36 + 4
= 4.
Since,
x^3 - 7x^2 + 6x + 4 = (x-6)* q(x) + 4
=) x^3 - 7x^2 + 6x + 4 - 4 = (x-6) * q(x)
=) x^3 - 7x^2 + 6x = (x-6) * q(x)
=) (x^3 - 7x^2 + 6x ) / (x-6) = q(x)
Hence q(x) = x^2 - x
Hope it's helpful to u.
According to Remainder theorem,
If a polynomial p(x) divided by x-a then remainder is equal to p(a)..
Let x^3 - 7x^2 + 6x + 4 as p(x) .
Hence value of a will be 6.
Hence remainder = p(6)
=) p(6) = x^3 - 7x^2 + 6x + 4
= 6^3 - 7* 6^2 + 6*6 + 4
= 216 - 252 + 36 + 4
= 4.
Since,
x^3 - 7x^2 + 6x + 4 = (x-6)* q(x) + 4
=) x^3 - 7x^2 + 6x + 4 - 4 = (x-6) * q(x)
=) x^3 - 7x^2 + 6x = (x-6) * q(x)
=) (x^3 - 7x^2 + 6x ) / (x-6) = q(x)
Hence q(x) = x^2 - x
Hope it's helpful to u.
micky161:
i hope in remainder theorem we do divide one & i think u are telling factor thorem
No ..I'm right..factor theorem states that if x-a is a factor of px then pa = 0..
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