1. Determine which of the following polynomials has ` \(x+1)\ ` a factor :
(i) ` \x^3+x^2+x+1\ `
(ii) ` \x^4+x^3+x^2+1\ `
(iii) ` \x^4+3x^3+3x^2+x+1\ `
(iv) ` \x^3-x^2-(2+\sqrt{2})x+\sqrt{2}\ `
solution :
(i) ` \x^3+x^2+x+1\ `
Let, ` \p(x)=x^3+x^2+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 ` \x^3+x^2+x+1\ `.
(ii) ` \x^4+x^3+x^2+1\ `
Let, ` \p(x)=x^4+x^3+x^2+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\ne0\ `
∴ By factor theorem, ` \x+1\ ` is not a factor of ` \x^4+x^3+x^2+1\ `.
(iii) ` \x^4+3x^3+3x^2+x+1\ `
Let, ` \p(x)=x^4+3x^3+3x^2+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\ `
` \=1ne0\ `
∴ By factor theorem, ` \x+1\ ` is not a factor of ` \x^4+3x^3+3x^2+x+1\ `.
(iv) ` \x^3-x^2-(2+\sqrt{2})x+\sqrt{2}\ `
Let, ` \x^3-x^2-(2+\sqrt{2})x+\sqrt{2}\ `
The zero of ` \(x+1)\ ` is
` \x+1=0\ `
` \x=-1\ `
So,
` \p(-1)\ `=` \(-1)^3-(-1)^2-(2+\sqrt{2})(-1)+\sqrt{2}\ `
` \=-1-1-(-2-\sqrt{2})+\sqrt{2}\ `
` \=-2+2+\sqrt{2}+\sqrt{2}\ `
` \=2\sqrt{2}ne0\ `
∴ By factor theorem, ` \x+1\ ` is not a factor of ` \x^3-x^2-(2+\sqrt{2})x+\sqrt{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)=2x^3+x^2-2x-1\ `,` \g(x)=x+1\ `
(ii) ` \p(x)=x^3+3x^2+3x+1\ `,` \g(x)=x+2\ `
(iii) ` \p(x)=x^3-4x^2+x+6\ `,` \g(x)=x-3\ `
solution :
(i) ` \p(x)=2x^3+x^2-2x-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)=x^3+3x^2+3x+1\ `,` \g(x)=x+2\ `
` \p(x)=x^3+3x^2+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\ `
` \= -1ne0\ `
∴ By factor theorem, ` \g(x)\ ` is not a factor of ` \p(x)\ `.
(iii) ` \p(x)=x^3-4x^2+x+6\ `,` \g(x)=x-3\ `
` \p(x)=x^3-4x^2+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)=x^2+x+k\ `
(ii) ` \p(x)=2x^2+kx+\sqrt{2}\ `
(iii) ` \p(x)=kx^2-\sqrt{2}x+1\ `
(iv) `\p(x)=kx^2-3x+k\ `
solution :
(i) ` \p(x)=x^2+x+k\ `
As ` \x-1\ ` is a factor of ` \p(x)=x^2+x+k\ `,
` \p(1)=0\ `
Now,
` \p(1)=1^2+1+k\ `
So,
` \1+1+k=0\ `
` \⇒2+k=0\ `
` \⇒k=-2\ `
So, the value of ` \k\ ` is ` \-2\ `.
(ii) ` \p(x)=2x^2+kx+\sqrt{2}\ `
As ` \x-1\ ` is a factor of ` \p(x)=2x^2+kx+\sqrt{2}\ `,
` \p(1)=0\ `
Now,
` \p(1)=2×1^2+k×1+\sqrt{2}\ `
So,
` \2+k+\sqrt{2}=0\ `
` \⇒k=-2-\sqrt{2}\ `
` \⇒k=-(2+\sqrt{2})\ `
So, the value of ` \k\ ` is ` \-(2+\sqrt{2})\ ` .
(iii) ` \p(x)=kx^2-\sqrt{2}x+1\ `
As `\x-1\ ` is a factor of ` \p(x)=kx^2-\sqrt{2}x+1\ `,
` \p(1)=0\ `
Now,
` \p(1)=k×1^2-\sqrt{2}×1+1\ `
So,
` \k-\sqrt{2}+1=0\ `
` \⇒k=\sqrt{2}-1\ `
So, the value of ` \k\ ` is ` \sqrt{2}-1\ `.
(iv) `\p(x)=kx^2-3x+k\ `
As ` \x+1\ ` is a factor of `\p(x)=kx^2-3x+k\ `,
` \p(1)=0\ `
Now,
` \p(1)=k×1^2-3×+k\ `
So,
` \k-3+k=0\ `
` \⇒2k=3\ `
` \⇒k=\frac{3}{2}\ `
So, the value of ` \k\ ` is ` \frac{3}{2}\ `.
4. Factorise :
(i) ` \12x^2+7x+3\ `
(ii) ` \2x^2+7x+3\ `
(iii) ` \6x^2+5x-6\ `
(iv) ` \3x^2-x-4\ `
solution :
(i) ` \12x^2+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,
` \12x^2-7x+1\ `
` \=12x^2+(-3+(-4))x+1\ `
` \=12x^2-3x-4x+1\ `
` \=3x(4x-1)-1(4x-1)\ `
` \(3x-1)(4x-1)\ `
The factor of ` \12x^2+7x+3\ ` is ` \(3x-1)(4x-1)\ `.
(ii) ` \2x^2+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,
` \2x^2+7x+3\ `
` \=2x^2+(1+6)x+3\ `
` \=2x^2+1x+6x+3\ `
` \=x(2x+1)+3(2x+1)\ `
` \=(x+3)(2x+1)\ `
The factor of ` \2x^2+7x+3\ ` is ` \(x+3)(2x+1)\ `.
(iii) ` \6x^2+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,
` \6x^2+5x-6\ `
` \=6x^2+(9+(-4))x-6\ `
` \=6x^2+9x-4x-6\ `
` \=3x(2x+3)-2(2x+3)\ `
` \(3x-2)(2x+3)\ `
The factor of ` \6x^2+5x-6\ ` is ` \(3x-2)(2x+3)\ `.
(iv) ` \3x^2-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,
` \3x^2-x-4\ `
` \=3x^2+(3+(-4))x-4\ `
` \=3x^2+3x-4x-4\ `
` \=3x(x+1)-4(x+1)\ `
` \(3x-4)(x+1)\ `
The factor of ` \3x^2-x-4\ ` is ` \(3x-4)(x+1)\ `.
5. Factorise :
(i) ` \x^3-2x^2-x+2\ `
(ii) ` \x^3-3x^2-9x-5\ `
(iii) ` \x^3+13x^2+32x+20\ `
(iv) ` \2y^3+y^2-2y-1\ `
solution :
(i) ` \x^3-2x^2-x+2\ `
Let,
` \p(x)=x^3-2x^2-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
` \x^3-2x^2-x+2\ `
` \=x^3-x^2-x^2+x-2x+2\ `
` \=x^2(x-1)-x(x-1)-2(x-1)\ `
` \=(x-1)(x^2-x-2)\ `
We could have also got this by dividing ` \p(x)\ ` by ` \x-1\ `.
Now ` \x^2-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:
` \x^2-x-2\ `
` \=x^2+1x-2x-2\ `
` \=x(x-1)-2(x+1)\ `
So,
` \x^3-2x^2-x+2\ `
` \=(x-1)(x-2)(x+1)\ `
(ii) ` \x^3-3x^2-9x-5\ `
Let,
` \p(x)=x^3-3x^2-9x-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
` \x^3-3x^2-9x-5\ `
` \=x^3+x^2-4x^2-4x-5x-5\ `
` \=x^2(x+1)-4x(x+1)-5(x+1)\ `
` \(x+1)(x^2-4x-5)\ `
We could have also got this by dividing ` \p(x)\ ` by ` \x+1\ `.
Now ` \x^2-4x-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:
` \x^2-4x-5\ `
` \=x^2+1x-5x-5\ `
` \=x(x+1)-5(x+1)\ `
` \=(x-5)(x+1)\ `
So,
` \x^3-3x^2-9x-5\ `
` \=(x+1)(x-5)(x+1)\ `
(iii) ` \x^3+13x^2+32x+20\ `
Let,
` \p(x)=x^3+13x^2+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
` \x^3+13x^2+32x+20\ `
` \=x^3+x^2+12x^2+12x+20x+20\ `
` \=x^2(x+1)+12x(x+1)+20(x+1)\ `
` \(x+1)(x^2+12x+20)\ `
We could have also got this by dividing ` \p(x)\ ` by ` \x+1\ `.
Now ` \x^2+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:
` \x^2+12x+20\ `
` \=x^2+2x+10x+20\ `
` \=x(x+2)+10(x+2)\ `
` \=(x+2)(x+10)\ `
So,
` \x^3+13x^2+32x+20\ `
` \=(x+1)(x+2)(x+10)\ `
(iv) ` \2y^3+y^2-2y-1\ `
Let,
` \p(x)=2y^3+y^2-2y-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
` \2y^3+y^2-2y-1\ `
` \=2y^3+2y^2-y^2-y-y-1\ `
` \=2y^2(y+1)-y(y+1)-1(y+1)\ `
` \=(y+1)(2y^2-y-1)\ `
We could have also got this by dividing ` \p(y)\ ` by ` \y-1\ `.
Now ` \2y^2-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:
` \2y^2-y-1\ `
` \=2y^2-2y+y-1\ `
` \=2y(y-1)+1(y-1)\ `
So,
` \2y^3+y^2-2Y-1\ `
` \=(y-1)(2y+1)(y+1)\ `
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