Question

Using factor theorem, factorize each of the following polynomials: (i) x3 – 6x2 + 3x + 10 (ii) 2y3 – 5y2 – 19y + 42

Answer 1

$\textbf{Given:}$

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

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

$\textbf{To find:}$

$\text{Factors of p(x) and q(y) }$

$\textbf{Solution:}$

$\text{Consider,}$

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

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

$p(-1)=-1-6-3+10=0$

$\text{By factor theorem,}$

$\textbf{ \bf(x+1) is a factor of p(x)}$

$\text{By using synthetic division,}$

$\begin{array}{r|cccc}-1&1&-6&3&10\\&-&-1&7&-10\\\cline{2-5}&1&-7&10&\boxed{0}\end{array}$

$x^2-7x+10$

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

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

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

$\therefore\,p(x)=(x+1)(x-5)(x-2)$

$\text{Consider,}$

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

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

$q(2)=16-20-38+42=0$

$\text{By factor theorem,}$

$\textbf{ \bf(y-2) is a factor of p(x)}$

$\text{By using synthetic division,}$

$\begin{array}{r|cccc}2&2&-5&-19&42\\&-&4&-2&-42\\\cline{2-5}&2&-1&-21&\boxed{0}\end{array}$

$2y^2-y-21$

$=2y^2-7y+6y-21$

$=y(2y-7)+3(2y-7)$

$=(y+3)(2y-7)$

$\therefore\bf\,q(y)=(y-2)(y+3)(2y-7)$

Find more:

Using factor theorem, factorize each of the following polynomial:

x³-10x²-53x-42

https://brainly.in/question/15904661

Factorise y2-5y+6 by using the factor theorem

https://brainly.in/question/3493349

$\text{}$

Answer 2

Answer:

GIVEN :

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

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

3

−6x

2

+3x+10

(ii) 2y^3 - 5y^2 -19y + 422y

3

−5y

2

−19y+42

TO FIND :

The factors the given polynomials.

SOLUTION :

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

3

−6x

2

+3x+10

(ii) 2y^3 - 5y^2 -19y + 422y

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 + 10x

3

−6x

2

+3x+10

Let f(x) be the given polynomial.

f(x)=x^3 - 6x^2 + 3x + 10f(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)+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=0x

2

−7x+10=0

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

2

−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)x

3

−6x

2

+3x+10=(x+1)(x−5)(x−2)

∴ the factors of the given polynomial x^3 - 6x^2 + 3x + 10x

3

−6x

2

+3x+10 are (x+1),(x-5) and (x-2)

(ii) 2y^3 - 5y^2 -19y + 422y

3

−5y

2

−19y+42

Let f(y) be the given polynomial.

f(y)=2y^3 - 5y^2 -19y + 42f(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)+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=02x

2

−11x+14=0

Now solve the quadratic equation by using the formula,

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

2

+bx+c=0

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

2a

−b±

b

2

−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}=

4

11±

121−112

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

4

11±

9

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

4

11±3

∴ x=\frac{11+3}{4}x=

4

11+3

and x=\frac{11-3}{4}x=

4

11−3

x=\frac{14}{4}x=

4

14

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

4

8

x=\frac{14}{4}x=

4

14

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

4

8

∴ x=\frac{7}{2}x=

2

7

and x=2 are the zeroes.

∴ x-\frac{7}{2}x−

2

7

and x-2 are the factors.

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

3

−5y

2

−19y+42=(x+3)(x−

2

7

)(x−2)

∴ the factors of the given polynomial 2y^3 - 5y^2 -19y + 422y

3

−5y

2

−19y+42 are (x+3),(x-\frac{7}{2}

2

7

) and (x-2).

Step-by-step explanation:

hope this will help you