Tags: Factor Theorem Examples and Solutions for Grade 9, Factor Theorem Worksheet PDF for Class IX, Factor Theorem Questions and Answers for ninth class, Revision Notes for Factor Theorem for 9^{th} Grade, Factor Theorem Practice Page
Factor Theorem
We know that if q(x) divides p(x) completely, that means p(x) is divisible by q(x) or, q(x) is a factor of p(x).
Now we will study a theorem which will help us to determine whether a polynomial q(x) is a factor of a polynomial p(x) or not without doing the actual division. This theorem is known as the factor theorem.
Factor Theorem: If p(x) is a polynomial of degree greater than 1 i.e. n > 1 and a is any real number such that p(a) = 0, then (x - a) is a factor of p(x) or we can say if (x - a) is a factor of p(x), then p(a) = 0
This implies,
(x + a) is a factor of a polynomial p(x) if p(-a) = 0.
(ax – b) is a factor of a polynomial p(x) if p(b/a) = 0.
(ax + b) is a factor of a polynomial p(x) if p(-b/a) = 0.
(x-a)(x-b) is a factor of a polynomial p(x) if p(a) =0 and p(b) = 0.
Example 1: Show that (x-1) is a factor of the polynomial x^{3} + 4x^{2} + x – 6.
Solution: Let p(x) = x^{3} + 4x^{2} + x – 6 be the given polynomial.
By factor theorem, (x-a) is a factor of a polynomial p(x) if p(a) = 0. Therefore, in order to prove that x – 1 is a factor of p(x), we have to show that p(1) = 0
Now,
P(x) = x^{3} + 4x^{2} + x – 6
P(1) = 1^{3} + 4 x 1^{2} + 1 – 6
1 + 4 + 1 – 6 = 0
Hence, (x-1) is a factor of p(x) = x^{3} + 4x^{2} + x – 6
Example 2: Find the remainder when 2x^{3}- 9x^{2}+ x+12 when divided by x - 1.
Solution: Let p(x) = 2x^{3}- 9x^{2}+ x+12 be the given polynomial.
Now,
P(x) = 2x^{3}- 9x^{2}+ x+12
P(1) = 2x1^{3} - 9 x (1)^{2} + 1 + 12
2 - 9 + 1 + 12 = 15 – 9 = 6
Therefore, 6 is the remainder when 2x^{3}- 9x^{2}+ x+12 is divided by x - 1.
Example 3: Without actual division prove that 4x^{4} – 2x^{3} - 6x^{2} + x + 1 is exactly divisible by 2x^{2} + x - 1.
Solution: Let f(x) = 4x^{4} – 2x^{3} - 6x^{2} + x + 1 and g(x) = 2x^{2} + x - 1 be the given polynomials.
We have, g(x) = 2x^{2} + x - 1 = 2x^{2} + 2x -1x - 1 = 2x(x+1)-1(x+1) = (2x-1) (x+1)
(2x-1) and (x+1) are factors of g(x)
In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that 2x-1 and x+1 are factors of f(x). For this it is sufficient to prove that f(1/2) = 0 and f(-1) = 0.
Now, f(x) = 4x^{4} – 2x^{3} - 6x^{2} + x + 1
f(1/2) = 4 x (1/2)^{4} – 2 x (1/2)^{3} - 6 x (1/2)^{2} + 1/2 + 1
= 1/4 – 1/4 – 3/2 + 1/2 + 1 = 0
f(-1) = 4 x (-1)^{4} – 2 x (-1)^{3} - 6 x (-1)^{2} - 1 + 1
= 4 + 2 - 6 – 1 + 1 = 0
(2x-1) and (x+1) are factors of f(x)
g(x) = (2x-1)(x+1) is a factors of f(x)
Hence, f(x) is exactly divisible by g(x)
Example 4: Find the value of a, if (x + a) is a factor of x^{3} + ax^{2} - 2x + a + 4.
Solution: Let p(x) = x^{3} + ax^{2} - 2x + a + 4 be the given polynomial. By factor theorem, (x + a) is a factor of p(x) if p(-a) = 0.
Now, p(-a) = 0
(-a)^{3} + a(-a)^{2} – 2(-a) + a + 4 = 0
-a^{3} + a^{3} + 2a + a + 4 = 0
3a + 4 = 0
a = -4/3
Hence, (x + a) is a factor of the given polynomial, if a = -4/3
Example 5: If ax^{3} + bx^{2} + x - 6 has x + 2 as a factor and leaves a remainder 4 when divided by (x-2), find the values of ‘a’ and ‘b’.
Solution: Let p(x) = ax^{3} + bx^{2} + x -6 be the given polynomial.
Now, (x+2) is a factor of p(x)
P(-2) = 0 (as x+2 = 0, x = -2)
a(-2)^{3} +b(-2)^{2 }+ (-2) – 6 = 0
-8a + 4b – 2 – 6= 0
-8a + 4b = 8
-2a + b = 2
It is given that p(x) leaves the remainder 4 when it is divided by (x-2).
Therefore, p(2) = 4 (as x – 2 = 0, x= -2)
a(2)^{3} + b(2)^{2} + 2 – 6 = 4
8a + 4b – 4 = 4
8a + 4b = 8
2a + b = 2
Adding (i) and (ii), we get
2b = 4
b = 2
Putting b = 2 in (i), we get
-2a + 2 = 2
-2a = 0
a = 0
Hence, a = 0 and b = 2
Example 6: If x^{2} – 1 is a factor of ax^{4} + bx^{3} + cx^{2} +dx + e, show that a + c + e = b + d = 0
Solution: Let p(x) = ax^{4} + bx^{3} + cx^{2} +dx + e be the given polynomial. Then,
x^{2} – 1 is a factor of p(x)
(x-1)(x+1) is a factor of p(x)
(x-1) and (x+1) are factors of p(x)
P(1) = 0 and p(-1) = 0 (as x – 1 = 0, x = 1 and x + 1 = 0, x = -1)
a + b + c + d + e = 0 and a – b + c – d + e = 0
Adding and subtracting these two equations, we get
2(a + c + e) = 0 and 2 (b + d) = 0
a + c + e = 0 and b + d = 0
a + c + e = b + d = 0
Example 7: What must be added to x^{3} - 3x^{2} – 12x + 19 so that the result is exactly divisible by x^{2} + x – 6.
Solution: Let p(x) = x^{3} - 3x^{2} – 12x + 19 and q(x) = x^{2} + x – 6. When p(x) is divided by q(x), the remainder is a linear expression in x. So, let r(x) = ax + b is added to p(x) so that p(x) + r(x) is divisible by q(x).
Let f(x) = p(x) + r(x). Then,
f(x) = x^{3} - 3x^{2} – 12x + 19 + ax + b
We have, q(x) = x^{2} + x – 6= x^{2} + 3x – 2x – 6 = x(x + 3) – 2(x + 3) = (x – 2) (x + 3)
Clearly, q(x) is divisible by (x – 2) and (x + 3) i.e. x – 2 and x + 3 are factors of q(x).
Therefore, f(x) will be divisible by q(x), if x – 2 and x + 3 are factors of f(x)
i.e., f(2) = 0 and f(-3) = 0
x^{3} - 3x^{2} – 12x + 19 + ax + b = 0
(2)^{3} – 3(2)^{2} – 12(2) + 19 + a(2) + b = 0
8 – 12 – 24 + 19 + 2a + b = 0
2a + b = 9 (i)
and x^{3} - 3x^{2} – 12x + 19 + ax + b = 0
(-3)^{3} – 3(-3)^{2} – 12(-3) + 19 + a(-3) + b = 0
-27 – 27 + 36 + 19 -3a + b = 0
1 – 3a + b = 0
-3a + b = -1 (ii)
Subtracting (i) from (ii) we get a = 2
Now substitute value of a = 2 in (i)
2 x 2 + b = 9
b = 5
Therefore, r(x) = ax + b
r(x) = 2x + 5
Hence, x^{3} - 3x^{2} – 12x + 19 will be divisible by x^{2} + x – 6, if 2x + 5 is added to it.
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