Question

plz solve question no. 5a​

Answer 1

Question :-- Factorise 3x³ - 17x² + 18x + 8 ?

Solution :----

Lets First assume this is Equal to 0.

3x³ - 17x² + 18x + 8 = 0

Splitting the terms and adding and subtracting 4x , we can write

→ 3x³ - 6x² - 11x² + 22x - 4x + 8 = 0

→ 3x²(x-2) - 11x(x-2) -4(x-2) = 0

Taking (x-2) common now,

→ (x-2)[ 3x² -11x -4 ] = 0

Now, if (x-2) = 0

x = 2

and, if 3x² - 11x -4 = 0

→ 3x² - 11x - 4 = 0

_____________________________

Now, Using Dharacharya formula For solving the Equation says that :---

Roots of Equation :- ax² + bx + c = 0 will be

===>> [ -b ± √b²-4ac / 2a ]

______________________________

we have Equation Now,

→ 3x² - 11x - 4 = 0

Here, a = 3, b = (-11) , c = (-4) ,

Putting all values now in Above Formula we get,

→ Roots are :--- [ 11 ± √{(-11)² - 4*3*(-4)}/2*3 ]

→ [ 11 ± √{121+48} / 6 }

→ [ 11 ± √169 /6 ]

→ [ 11 ± 13 / 6 ]

or, we can write,

→ Roots will be :- (11+13)/6 and (11-13/6)

→ Required Roots :- 24/6 and (-2)/6

→ Rest Two Roots = 3 and (-1/3)

______________________________

Hence, all three Roots of Given Equation will be 2 , 3 and (-1/3) ....

Answer 2

❏ Question:-

@ Factories it , 3x³-17x²+18x+8

❏ Solution:-

To solve this kind of polynomial of degree 3.

we will use here Vanishing Method or VM.

[The method is discussed in detail at below

please go through it]

Let, f(x)=3x³-17x²+18x+8

❍ Now, 1st part of the problem:-

✦ Now we will try with first x=0.

➔ 3x³-17x²+18x+8 = 3(0)³-17(0)²+18(0)+8

= 0-0+0+8

= 8 ≠ 0

✦ Now we will try with first x=1.

➔ 3x³-17x²+18x+8 = 3(1)³-17(1)²+18(1)+8

= 3-17+18+8

= 12 ≠ 0

✦ Now we will try with first x=-1.

➔ 3x³-17x²+18x+8 = 3(-1)³-17(-1)²+18(-1)+8

= -3+17-18+8

= +14-10

= 4 ≠ 0

✦ Now we will try with first x= 2.

➔ 3x³-17x²+18x+8 = 3(2)³-17(2)²+18(2)+8

= 24-68+36+8

= -44+44

= 0

So, by recalling Remainder Theorm we can say that 2 is a root of this polynomial.

∴ x=2

➝ (x-2)=0

So, we can say that (x-2) is a factor of the polynomial f(x).

━━━━━━━━━━━━━━━━━━━━━━━

❍ Now, 2nd part of solving:-

➔ 3x³-17x²+18x+8

➔ 3x³-6x²-11x²+22x-4x+8

➔ 3x²(x-2)-11x(x-2)-4(x-2)

➔ (x-2)(3x²-11x-4)

[now, applying middle term method to the 2nd part]

➔ (x-2)[3x²-12x+x-4]

➔ (x-2)[3x(x-4)+1(x-4)]

➔ (x-2)[(x-4)(3x+1)]

➔ (x-2)(x-4)(3x+1)

∴ 3x³-17x²+18x+8 = (x-2) (x-4) (3x+1) (Ans).

━━━━━━━━━━━━━━━━━━━━━━━

✰ Extra activities:-

So, from the above we can see that

(x-2) , (x-4) and (3x+1) are the factors of the polynomial f(x)=3x³-17x²+18x+8.

Hence, The roots are

• (x-2)=0

=> x=2

• (x-4)=0

=> x=4

• (3x+1)=0

=> x = $\sf\bf\frac{-1}{3}$

━━━━━━━━━━━━━━━━━━━━━━━

❏ Theorem and Formula applied:--

━━━━━━━━━━━━━━━━━━━━━━━

(1) Vanishing Method:-

=> This method is applicable to determine the factorisation of a polynomial of degree more than 2 . i.e, degree > 2.

In this method , actually we takes different values of the variable (of the polynomial) so that the expression vanishes or 0.

Basically, we use x= -3, -2, -1, 0, 1, 2, 3 etc.

Now, the value for which the expression becomes 0 , that value is taken under consideration .

ex:- suppose for the value of variable X=1 the expression vanishes,then we will take (X-1) as a factor of that particular polynomial.

Or,

(2) Remainder Theorem:-

For a value of X = a , if the polynomial f(X) becomes zero(0), i.e, f(X=a). then the (x-a) is the factor of that polynomial f(X).

N.B:- Here you can see , VM or R.T are same

almost .

━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

** Please Mark it if it helps u:-