Question

9. Using factor theorem, factorize each of the following polynomials: (i) x 3 – 6x2 + 3x + 10 (ii) 2y3 – 5y2 – 19y + 42

Answer 1

GIVEN :

Using factor theorem, factorize each of the following given polynomials :

(i) $x^3 - 6x^2 + 3x + 10$

(ii) $2y^3 - 5y^2 -19y + 42$

TO FIND :

The factors the given polynomials.

SOLUTION :

Given that the polynomials (i) $x^3 - 6x^2 + 3x + 10$

(ii) $2y^3 - 5y^2 -19y + 42$

Now factorise the given polynomials by using Factor theorem.

Factor theorem states that :

A polynomial f(x) has a factor in the form of (x-a) if and only if  the value of f(a)=0.

(i) $x^3 - 6x^2 + 3x + 10$

Let f(x) be the given polynomial.

$f(x)=x^3 - 6x^2 + 3x + 10$

Put x=-1 in the above polynomial we get,

$f(-1)=(-1)^3-6(-1)^2+3(-1)+10$

$=-1-6-3+10$

$=0$

∴ f(-1)=0 and -1 is a zero.

∴ x+1 is a factor of the given polynomial.

By using Synthetic division we can find the factors.

-1_| 1     -6       3        10

     0     -1       7       -10

   _________________

     1     -7       10        0

 

$x^2-7x+10=0$

$x^2-5x-2x+10=0$

$x(x-5)-2(x-5)=0$

$(x-5)(x-2)=0$

x-5=0 or x-2=0

∴ x=5 , 2 are the zeroes.

∴ $x^3 - 6x^2 + 3x + 10=(x+1)(x-5)(x-2)$

∴ the factors of the given polynomial $x^3 - 6x^2 + 3x + 10$ are (x+1),(x-5) and (x-2)

(ii) $2y^3 - 5y^2 -19y + 42$

Let f(y) be the given polynomial.

$f(y)=2y^3 - 5y^2 -19y + 42$

Put y=-3 in the above polynomial we get,

$f(-3)=2(-3)^3-5(-3)^2-19(-3)+42$

$=-54-45+57+42$

By adding the like terms,

$=0$

∴ f(-3)=0 and -3 is a zero.

∴ x+3 is a factor of the given polynomial.

By using Synthetic division we can find the factors.

-3_|  2    -5      -19        42

       0     -6      33       -42

     _________________

       2     -11      14         0

$2x^2-11x+14=0$

Now solve the quadratic equation by using the formula,

For a quadratic equation $ax^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here a=2 , b=-11 and c=14

Substitute the values in the formula we get

$x=\frac{-(-11)\pm \sqrt{(-11)^2-4(2)(14)}}{2(2)}$

$=\frac{11\pm \sqrt{121-112}}{4}$

$=\frac{11\pm \sqrt{9}}{4}$

$=\frac{11\pm 3}{4}$

∴ $x=\frac{11+3}{4}$ and $x=\frac{11-3}{4}$

$x=\frac{14}{4}$ and $x=\frac{8}{4}$

$x=\frac{14}{4}$ and $x=\frac{8}{4}$

∴ $x=\frac{7}{2}$ and x=2 are the zeroes.

∴ $x-\frac{7}{2}$ and x-2 are the factors.

∴ $2y^3 - 5y^2 -19y + 42=(x+3)(x-\frac{7}{2})(x-2)$

∴ the factors of the given polynomial $2y^3 - 5y^2 -19y + 42$ are (x+3),(x-$\frac{7}{2}$) and (x-2).

Answer 2

Answer:

Using factor theorem, factorize each of the following given polynomials :

(i) x^3 - 6x^2 + 3x + 10x3−6x2+3x+10

(ii) 2y^3 - 5y^2 -19y + 422y3−5y2−19y+42

TO FIND :

The factors the given polynomials.

SOLUTION :

Given that the polynomials (i) x^3 - 6x^2 + 3x + 10x3−6x2+3x+10

(ii) 2y^3 - 5y^2 -19y + 422y3−5y2−19y+42

Now factorise the given polynomials by using Factor theorem.

Factor theorem states that :

A polynomial f(x) has a factor in the form of (x-a) if and only if  the value of f(a)=0.

(i) x^3 - 6x^2 + 3x + 10x3−6x2+3x+10

Let f(x) be the given polynomial.

f(x)=x^3 - 6x^2 + 3x + 10f(x)=x3−6x2+3x+10

Put x=-1 in the above polynomial we get,

f(-1)=(-1)^3-6(-1)^2+3(-1)+10f(−1)=(−1)3−6(−1)2+3(−1)+10

=-1-6-3+10=−1−6−3+10

=0=0

∴ f(-1)=0 and -1 is a zero.

∴ x+1 is a factor of the given polynomial.

By using Synthetic division we can find the factors.

-1_| 1     -6       3        10

     0     -1       7       -10

   _________________

     1     -7       10        0

 

x^2-7x+10=0x2−7x+10=0

x^2-5x-2x+10=0x2−5x−2x+10=0

x(x-5)-2(x-5)=0x(x−5)−2(x−5)=0

(x-5)(x-2)=0(x−5)(x−2)=0

x-5=0 or x-2=0

∴ x=5 , 2 are the zeroes.

∴ x^3 - 6x^2 + 3x + 10=(x+1)(x-5)(x-2)x3−6x2+3x+10=(x+1)(x−5)(x−2)

∴ the factors of the given polynomial x^3 - 6x^2 + 3x + 10x3−6x2+3x+10 are (x+1),(x-5) and (x-2)

(ii) 2y^3 - 5y^2 -19y + 422y3−5y2−19y+42

Let f(y) be the given polynomial.

f(y)=2y^3 - 5y^2 -19y + 42f(y)=2y3−5y2−19y+42

Put y=-3 in the above polynomial we get,

f(-3)=2(-3)^3-5(-3)^2-19(-3)+42f(−3)=2(−3)3−5(−3)2−19(−3)+42

=-54-45+57+42=−54−45+57+42

By adding the like terms,

=0=0

∴ f(-3)=0 and -3 is a zero.

∴ x+3 is a factor of the given polynomial.

By using Synthetic division we can find the factors.

-3_|  2    -5      -19        42

       0     -6      33       -42

     _________________

       2     -11      14         0

2x^2-11x+14=02x2−11x+14=0

Now solve the quadratic equation by using the formula,

For a quadratic equation ax^2+bx+c=0ax2+bx+c=0

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac

Here a=2 , b=-11 and c=14

Substitute the values in the formula we get

x=\frac{-(-11)\pm \sqrt{(-11)^2-4(2)(14)}}{2(2)}x=2(2)−(−11)±(−11)2−4(2)(14)

=\frac{11\pm \sqrt{121-112}}{4}=411±121−112

=\frac{11\pm \sqrt{9}}{4}=4