FACTOR THEROEM PRACTICE QUESTIONS

Problem 1 :

Determine whether (x+1) is a factor of the following polynomials.

(i)  6x⁴+7x³-5x-4

(ii)  2x⁴+9x³+2x²+10x+15

(iii)  3x³+8x²-6x-5

(iv)  x³-14x²+3x+12

Problem 2 :

Determine whether (x+4) is a factor of

x³ + 3x² - 5x + 36

Problem 3 :

Using factor theorem show that (x-1) is a factor of  

4x³-6x²+9x-7

Problem 4 :

Determine whether (2x+1) is a factor of

4x³+4x²-x-1

Problem 5 :

Determine the value of p if (x+3) is a factor of

x³-3x²-px+24

(i)  Solution :

6x⁴+7x³-5x-4

By factor theorem, if p(-1)  =  0, then (x+1) is a factor of

p(x) =  6x⁴+7x³-5x-4

p(-1)  = 6(-1)⁴+7(-1)³-5(-1)-4

=  6-7+5-4

p(-1)  =  0

(x+1) is a factor of the given polynomial.

(ii)  Solution :

2x⁴+9x³+2x²+10x+15

By factor theorem, if p(-1) = 0, then (x+1) is a factor of 

p(x)  =  2x⁴+9x³+2x²+10x+15

p(-1)  =  2(-1)⁴+9(-1)³+2(-1)²+10(-1)+15

=  2-9+2-10+15

=  0

So, (x+1) is a factor of the given polynomial.

(iii)  Solution :

3x³+8x²-6x-5

By factor theorem, if p(-1) = 0, then (x+1) is a factor of 

p(x)  =  3x³+8x²-6x-5

p(-1)  =  3(-1)³+8(-1)²-6(-1)-5

=  -3+8+6-5

p(-1)   ≠  0

(x+1) ix not the factor of 3x³+8x²-6x-5.

(iv)  Solution :

x³-14x²+3x+12

By factor theorem, if p(-1) = 0, then (x+1) is a factor of

p(x)  =  x³-14x²+3x+12

p(-1)  =  (-1)³-14(-1)²+3(-1)+12

p(-1)  =  -1-14-3+12

p(-1)  ≠  0

(x+1) is not the factor of x³-14x²+3x+12.

(2)  Solution :

By factor theorem, if p(-4) = 0, then (x+4) is a factor of

p(x)  =  x³ + 3x² - 5x + 36

p(-4)  =  (-4)³+3(-4)²-5(-4)+36

=  -64+48+20+36

p(-4)  ≠  0

(x+4) is not the factor of x³+3x²-5x+36.

(3)  Solution :

By factor theorem, if p(1) = 0, then (x-1) is a factor of 

p(x)  =  4x³-6x²+9x-7

p(1)  =  4(1)³-6(1)²+9(1)-7

=  4-6+9-7

p(1)  =  0

(x-1) is the factor of 4x³-6x²+9x-7.

(4)  Solution :

By factor theorem, if p(-1/2) = 0, then (2x+1) is a

Factor of p(x)  =  4x³+4x²-x-1

p(-1/2)  =  4(-1/2)³+4(-1/2)²-(-1/2)-1

=  -1/2+1+1/2-1

p(-1/2)  =  0

(2x+1) is the factor of 4x³+4x²-x-1.

(5)  Solution :

By factor theorem, if p(-3) = 0, then (x+3) is a 

factor of p(x)  =  x³-3x²-px+24

p(-3)  =  (-3)³-3(-3)²- p(-3)+24

This implies that  -27-27+3p+24  =  0

-30 + 3p  =  0

3p  =  30

p  =  10

So, the value of p is 10.

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