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REMAINDER
THEOREM
Review:
Divide using synthetic division to find the remainder.
1. p(x) = x3 – 7x – 6, D(x) = x - 4
2. p(x) = 3x2 – 4x + 7, D(x) = x - 1
3. p(x) = 3x3 – 2x2 - 7x + 6, D(x) = x + 1
4. p(x) = 2x3 – 5 - 7, D(x) = x - 2
5. p(x) = x3 + 5 , D(x) = x + 2
The Remainder Theorem
When we divide a
polynomial f(x) by x-c the
remainder r equals f(c)
The remainder theorem is based on
synthetic division, which is the
process of dividing a polynomial f(x)
by a polynomial D(x) and finding
the remainder. This is written as ,
where f(x) is the dividend, Q(x) is the
quotient, D(x) is the divisor, and R(x)
is the remainder.
When we divide a
polynomial f(x) by x-
c we get:
f(x) = (x-c)·q(x) + r(x)
But r(x) is simply the
constant r (remember? when we
divide by (x-c) the remainder is a
constant) .... so we get this:
f(x) = (x-c)·q(x) + r
Now see what happens when
we have x equal to c:
f(c) = (c-c)·q(c) + r
f(c) = (0)·q(c) + r
f(c) = r
Illustrative Examples:
1. 2x2 - 5x - 1 divided by x - 3
Equate x – 3 = 0
x = 3
+ 3
Replace all x = 3
= 2(3)2 – 5(3) - 1
= 2(9) – 5(3) - 1
= 18 – 15 - 1
= 2
2x2 - 5x - 1
2. 2x2 - 5x - 1 divided by x + 4
F(c) = -4
x + 4 = 0 - 4
= 2(-4)2 – 5(-4) - 1
= 2(16) – 5(-4) - 1
= 32 – (-20) - 1
= 51
3. f (x) = −x3 + 6x − 7 at x = 2
= -3
4. f (x) = x3 + x2 − 5x − 6 at x = 2
= -4
5. f (x) = x5 − 47x3 − 16x2 + 8x + 52
at x = 7
= 10

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Remainder theorem

  • 2. Review: Divide using synthetic division to find the remainder. 1. p(x) = x3 – 7x – 6, D(x) = x - 4
  • 3. 2. p(x) = 3x2 – 4x + 7, D(x) = x - 1
  • 4. 3. p(x) = 3x3 – 2x2 - 7x + 6, D(x) = x + 1
  • 5. 4. p(x) = 2x3 – 5 - 7, D(x) = x - 2
  • 6. 5. p(x) = x3 + 5 , D(x) = x + 2
  • 7. The Remainder Theorem When we divide a polynomial f(x) by x-c the remainder r equals f(c)
  • 8. The remainder theorem is based on synthetic division, which is the process of dividing a polynomial f(x) by a polynomial D(x) and finding the remainder. This is written as , where f(x) is the dividend, Q(x) is the quotient, D(x) is the divisor, and R(x) is the remainder.
  • 9. When we divide a polynomial f(x) by x- c we get: f(x) = (x-c)·q(x) + r(x)
  • 10. But r(x) is simply the constant r (remember? when we divide by (x-c) the remainder is a constant) .... so we get this: f(x) = (x-c)·q(x) + r
  • 11. Now see what happens when we have x equal to c: f(c) = (c-c)·q(c) + r f(c) = (0)·q(c) + r f(c) = r
  • 12. Illustrative Examples: 1. 2x2 - 5x - 1 divided by x - 3 Equate x – 3 = 0 x = 3 + 3
  • 13. Replace all x = 3 = 2(3)2 – 5(3) - 1 = 2(9) – 5(3) - 1 = 18 – 15 - 1 = 2 2x2 - 5x - 1
  • 14. 2. 2x2 - 5x - 1 divided by x + 4 F(c) = -4 x + 4 = 0 - 4 = 2(-4)2 – 5(-4) - 1 = 2(16) – 5(-4) - 1
  • 15. = 32 – (-20) - 1 = 51 3. f (x) = −x3 + 6x − 7 at x = 2 = -3
  • 16. 4. f (x) = x3 + x2 − 5x − 6 at x = 2 = -4
  • 17. 5. f (x) = x5 − 47x3 − 16x2 + 8x + 52 at x = 7 = 10