 |
| Factor Theorem |
|
|
| If p(x), a polynomial in x is divided by x-a and the remainder = f (a) is zero, then (x-a) is a factor of p(x). |
| |
|
| |
| When p(x) is divided by x-a, |
| |
| R = p(a) (by remainder theorem) |
| |
| p(x) = (x-a).q(x)+p(a) |
| |
| (Dividend = Divisor x quotient + Remainder Division Algorithm) |
| |
| But p(a) = 0 is given. |
| |
| Hence p(x) = (x-a).q(x) |
| |
 |
| |
| Conversely if x-a is a factor of p(x) then p(a)=0. |
| |
| p(x) = (x-a).q(x) + R |
| |
| If (x-a) is a factor then the remainder should be zero (x-a divides p(x) |
| |
| exactly) |
| |
| R=0 |
| |
| By remainder theorem, R = p(a) |
| |
 p(a)=0 |
| |
| Illustrative Examples |
| |
| Example 1: |
| |
| Determine whether x-2 is a factor of x2-7x+10. |
| |
| Suggested answer: |
| |
| p(x) = x2-7x+10 is divided by x-2. |
| |
| R = p(2) |
| |
| = 4-14+10=0, R=0 |
| |
 |
| |
| Example 2: |
| |
| Determine whether x-3 is a factor of x3-3x2+4x-12. |
| |
| Suggested answer: |
| |
| p(x) = x3-3x2+4x-12 is divided by x-3. |
| |
| R = p(3) |
| |
| = 33- 3 x 32 + 4 x 3-12 |
| |
| = 27-27+12-12=0 |
| |
 |
| |
| Example 3: |
| |
| Show that x+1 is a factor of 2x3+5x2-9x-12. |
| |
| Suggested answer: |
| |
| Let p(x) = 2x3+5x2-9x-12 |
| |
when p(x) is  by x+1 [x- (-1)], |
| |
| R = p(-1) |
| |
| = 2(-1)3+5(-1)2-9(-1)-12 |
| |
| = -2+5+9-12 |
| |
| = -14+14 = 0 |
| |
| R = 0 |
| |
 |
| |
| Example 4: |
| |
| Find a so that x4+2x3-ax2+x-2 has (x+2) as its factor. |
| |
| Suggested answer: |
| |
| Let p(x) = x4+2x3-ax2+x-2 |
| |
when p(x) is  by x+2 [x- (-2)], |
| |
| R = p(-2) |
| |
| = (-2)4+2(-2)3-a(-2)2-2-2 |
| |
| = 16-16-4a-4 |
| |
| = -4a - 4 |
| |
| As (x+2) is a factor, R=0 |
| |
| -4a - 4 = 0 |
| |
| a = -1 |
| |
| Example 5: |
| |
| If (x-2) and (x-3) are factors of x3+ax2+bx+12, find a and b. |
| |
| Suggested answer: |
| |
| Let p(x) = x3+ax2+bx+12 |
| |
when p(x) is  x-2, |
| |
| |
| |
| R = p(2) |
| |
| = 23+a(2)2+b(2)+12 |
| |
| = 8+4a+2b+12 |
| |
| = 20+4a+2b |
| |
| But x-2 is a factor, |
| |
| 20+4a+2b = 0 |
| |
 |
| |
 |
| |
when p(x) is  by x-3, |
| |
| R = p(3) |
| |
| = 27+9a+3b+12 |
| |
| = 39+9a+3b |
| |
| But x-3 is a factor, |
| |
 |
| |
| 39+9a+3b=0 |
| |
[9a+3b=-39]  3 |
| |
| 3a+b = -13 …(2) |
| |
| Subtracting equation (1) from (2), we get, |
| |
| 3a+b-2a-b = -13+10 |
| |
| |
| |
| a = -3 |
| |
| Substitute a = -3 in equation (1), |
| |
| 2a+b = -10 |
| |
| -6+b = -10 |
| |
| b = -4 |
| |
| Example 6: |
| |
| Factorise using factor theorem. |
| |
| a) 2x2+x-3 |
| |
| b) 5x2+6x-8 |
| |
| Suggested answer: |
| |
| a) 2x2+x-3 |
| |
| Let p(x) = 2x2+x-3 |
| |
| The factors of the constant 3 are +1, -1, +3, -3. |
| |
| Let us find p(1) = 2+1-3 = 0 |
| |
 |
| |
| Divide 2x2+x-3 by x-1 to get the other factor. |
| |
 |
| |
| Hence 2x2+x-3 = (2x+3) (x-1) . |
| |
| b) 5x2+6x-8 |
| |
| Suggested answer: |
| |
| Let p(x) = 5x2+6x-8 |
| |
| Factors of the constant '8' are 1,-1,2,-2,4, -4 and 8,-8. |
| |
| Let us find the value of p(x) at each one of them to get one factor. |
| |
 |
| |
| p(-2) = 20-12-8 = 0 |
| |
 |
| |
| Divide 5x2+6x-8 by x+2 to get the other factor. |
| |
 |
| |
 |
| |
