1. Determine which of the following polynomials has (x+1) a factor :
(i) x3+x2+x+1
(ii) x4+x3+x2+1
(iii) x4+3x3+3x2+x+1
(iv) x3-x2-(2+√2)x+√2
solution :
(i) x3+x2+x+1
Let, p(x)=x3+x2+x+1
The zero of (x+1) is
x+1=0
x=-1
So,
p(-1)=(-1)3+(-1)2+(-1)+1
=-1+1-1+1
=0
∴ By factor theorem, x+1 is a factor of x3+x2+x+1.
(ii) x4+x3+x2+1
Let, p(x)=x4+x3+x2+1
The zero of (x+1) is
x+1=0
x=-1
So,
p(-1)=(-1)4+(-1)3+(-1)2+(-1)+1
=1-1+1-1+1
=1≠0
∴ By factor theorem, x+1 is not a factor of x4+x3+x2+1.
(iii) x4+3x3+3x2+x+1
Let, p(x)=x4+3x3+3x2+x+1
The zero of (x+1) is
x+1=0
x=-1
So,
p(-1)=(-1)4+3×(-1)3+3×(-1)2+(-1)+1
=1-3+3-1+1
=1≠0
∴ By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1.
(iv) x3-x2-(2+√2)x+√2
Let, x3-x2-(2+√2)x+√2
The zero of (x+1) is
x+1=0
x=-1
So,
p(-1)=(-1)3-(-1)2-(2+√2)(-1)+√2
=-1-1-(-2-√2)+√2
=-2+2+√2+√2
=2√2≠0
∴ By factor theorem, x+1 is not a factor of x3-x2-(2+√2)x+√2.
2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x)=2x3+x2-2x-1,g(x)=x+1
(ii) p(x)=x3+3x2+3x+1,g(x)=x+2
(iii) p(x)=x3-4x2+x+6,g(x)=x-3
solution :
(i) p(x)=2x3+x2-2x-1,g(x)=x+1
g(x)=0
⇒x+1=0
⇒x=-1
The zero of g(x) is -1
Now,
p(-1)=2(-1)3+(-1)2-2(-1)-1
=-2+1+2-1
=-1+1
=0
∴ By factor theorem, g(x) is a factor of p(x).
(ii) p(x)=x3+3x2+3x+1,g(x)=x+2
p(x)=x3+3x2+3x+1,g(x)=x+2
g(x)=0
⇒x+2=0
x=-2
The zero of g(x) is − 2
Now,
p(-2)=(-2)3+3(-2)2+3(-2)+1
=-8+12-6+1
=-1≠0
∴ By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x3-4x2+x+6,g(x)=x-3
p(x)=x3-4x2+x+6,g(x)=x-3
g(x)=0
⇒x-3=0
⇒x=3
The zero of g(x) is − 1.
Now,
p(3)=(3)3-4(3)2+(3)+6
=27-36+3+6
=0
∴ By factor theorem, g(x) is a factor of p(x).
3. Find the value of k, if x-1 is a factor of p(x) in each of the following cases:
(i) p(x)=x2+x+k
(ii) p(x)=2x2+kx+√2
(iii) p(x)=kx2-√2x+1
(iv) p(x)=kx2-3x+k
solution :
(i) p(x)=x2+x+k
As x-1 is a factor of p(x)=x2+x+k,
p(1)=0
Now,
p(1)=12+1+k
So,
1+1+k=0
⇒2+k=0
⇒k=-2
So, the value of k is -2.
(ii) p(x)=2x2+kx+√2
As x-1 is a factor of p(x)=2x2+kx+√2,
p(1)=0
Now,
p(1)=2×12+k×1+√2
So,
2+k+√2=0
⇒k=-2-√2
⇒k=-(2+√2)
So, the value of k is -(2+√2) .
(iii) p(x)=kx2-√2x+1
As x-1 is a factor of p(x)=kx2-√2x+1,
p(1)=0
Now,
p(1)=k×12-√2×1+1
So,
k-√2+1=0
⇒k=√2-1
So, the value of k is √2-1.
(iv) p(x)=kx2-3x+k
As x+1 is a factor of p(x)=kx2-3x+k,
p(1)=0
Now,
p(1)=k×12-3×+k
So,
k-3+k=0
⇒2k=3
⇒k=32
So, the value of k is 32.
4. Factorise :
(i) 12x2+7x+3
(ii) 2x2+7x+3
(iii) 6x2+5x-6
(iv) 3x2-x-4
solution :
(i) 12x2+7x+3
By splitting method:
If we can find two numbers p and q such that,
p+q=-7 and pq=12×1=12,
then we can get the factor.
So, let us look for the pair of factors of 12.
Some are 1 and 12, 2 and 6, − 3 and − 4, 3 and 4 etc.
of these pairs, − 3 and − 4 will giue us p+q=-7.
So,
12x2-7x+1
=12x2+(-3+(-4))x+1
=12x2-3x-4x+1
=3x(4x-1)-1(4x-1)
(3x-1)(4x-1)
The factor of 12x2+7x+3 is (3x-1)(4x-1).
(ii) 2x2+7x+3
By splitting method:
If we can find two numbers p and q such that,
p+q=7 and pq=3×2=6,
then we can get the factor.
So, let us look for the pair of factors of 6.
Some are 1 and 6, 2 and 3, − 3 and − 2 etc.
of these pairs, 1 and 6 will giue us p+q=7.
So,
2x2+7x+3
=2x2+(1+6)x+3
=2x2+1x+6x+3
=x(2x+1)+3(2x+1)
=(x+3)(2x+1)
The factor of 2x2+7x+3 is (x+3)(2x+1).
(iii) 6x2+5x-6
By splitting method:
If we can find two numbers p and q such that,
p+q=5 and pq=6×(-6)=-36,
then we can get the factor.
So, let us look for the pair of factors of − 36.
Some are 1 and − 36, 6 and − 6, − 9 and − 4, 3 and − 12 etc.
of these pairs, 9 and − 4 will giue us p+q=-5.
So,
6x2+5x-6
=6x2+(9+(-4))x-6
=6x2+9x-4x-6
=3x(2x+3)-2(2x+3)
(3x-2)(2x+3)
The factor of 6x2+5x-6 is (3x-2)(2x+3).
(iv) 3x2-x-4
By splitting method:
If we can find two numbers p and q such that,
p+q=-1 and pq=3×(-4)=-12,
then we can get the factor.
So, let us look for the pair of factors of − 12.
Some are 1 and − 12, 2 and − 6, − 3 and 4, 3 and − 4 etc.
of these pairs, 3 and − 4 will giue us p+q=-1.
So,
3x2-x-4
=3x2+(3+(-4))x-4
=3x2+3x-4x-4
=3x(x+1)-4(x+1)
(3x-4)(x+1)
The factor of 3x2-x-4 is (3x-4)(x+1).
5. Factorise :
(i) x3-2x2-x+2
(ii) x3-3x2-9x-5
(iii) x3+13x2+32x+20
(iv) 2y3+y2-2y-1
solution :
(i) x3-2x2-x+2
Let,
p(x)=x3-2x2-x+2
We shall now look for all the factors of 2.
Some of these are ± 1, ± 2.
By trial, we find that p(1)=0.
So,
x-1 is a factor of p(x).
Now we see that
x3-2x2-x+2
=x3-x2-x2+x-2x+2
=x2(x-1)-x(x-1)-2(x-1)
=(x-1)(x2-x-2)
We could have also got this by dividing p(x) by x-1.
Now x2-x-2 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
By splitting the middle term, we have:
x2-x-2
=x2+1x-2x-2
=x(x-1)-2(x+1)
So,
x3-2x2-x+2
=(x-1)(x-2)(x+1)
(ii) x3-3x2-9x-5
Let,
p(x)=x3-3x2-9x-5
We shall now look for all the factors of − 5.
Some of these are ± 1, ± 5.
By trial, we find that p ( − 1 ) = 0.
So, x+1 is a factor of p(x).
Now we see that
x3-3x2-9x-5
=x3+x2-4x2-4x-5x-5
=x2(x+1)-4x(x+1)-5(x+1)
(x+1)(x2-4x-5)
We could have also got this by dividing p(x) by x+1.
Now x2-4x-5 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
By splitting the middle term, we have:
x2-4x-5
=x2+1x-5x-5
=x(x+1)-5(x+1)
=(x-5)(x+1)
So,
x3-3x2-9x-5
=(x+1)(x-5)(x+1)
(iii) x3+13x2+32x+20
Let,
p(x)=x3+13x2+32x+20
We shall now look for all the factors of 20.
Some of these are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20.
By trial, we find that p(-1)=0.
So, x+1 is a factor of p(x).
Now we see that
x3+13x2+32x+20
=x3+x2+12x2+12x+20x+20
=x2(x+1)+12x(x+1)+20(x+1)
(x+1)(x2+12x+20)
We could have also got this by dividing p(x) by x+1.
Now x2+12x+20 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
By splitting the middle term, we have:
x2+12x+20
=x2+2x+10x+20
=x(x+2)+10(x+2)
=(x+2)(x+10)
So,
x3+13x2+32x+20
=(x+1)(x+2)(x+10)
(iv) 2y3+y2-2y-1
Let,
p(x)=2y3+y2-2y-1
We shall now look for all the factors of − 1.
Some of these are ± 1.
By trial, we find that p(1)=0.
So, y-1 is a factor of p(y).
Now we see that
2y3+y2-2y-1
=2y3+2y2-y2-y-y-1
=2y2(y+1)-y(y+1)-1(y+1)
=(y+1)(2y2-y-1)
We could have also got this by dividing p(y) by y-1.
Now 2y2-y-1 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
By splitting the middle term, we have:
2y2-y-1
=2y2-2y+y-1
=2y(y-1)+1(y-1)
So,
2y3+y2-2Y-1
=(y-1)(2y+1)(y+1)
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