Question

Find the zeros
of the polynomial 3x²-10x+8​

Answer 1

Answer:

here is your answer- 36¾

Answer 2

Concept used -

=> Here, we use splitting the middle term to find the zeroes of this polynomial.

=> First, we assume the following -

• 3x²-10x+8​ = 0

=> This gives us a quadratic equation, in which the highest power = 2 and 2 roots are possible.

=> Now, we split the middle term, find the factors, and finally find the zeroes of this polynomial.

Splitting the middle term -

=> In order to split the middle term, i.e. -10x we have to use the sum-product rule -

• Here, we must split -10 into two parts such that the sum of the two parts is -10 and the product of the two parts is 3 × 8 => 24.

=> We know that +24 can be expressed in the following ways -

• -6 × -4

• 6 × 4
• 12 × 2
• -12 × -2
• 8 × 3
• -8 × -3

=> Out of these pairs, it is clear that -6 × -4 gives product +24 and has sum

( -6 -4 = -10) equal to -10 .

=> Thus, the required parts will be -6x and -4x

Now, we have -

=> 3x²-10x+8​ = 0 [Taking -10x = -6x -4x]

=> 3x²-6x -4x +8​ = 0 [Taking out a common factors 3x and -4]

=> 3x(x-2) - 4(x-2) = 0 [Factorizing]

=> (3x-4)(x-2) = 0 [Now, x could be ]

=> (3x-4) = 0 or (x-2) = 0

=> 3x = 4 or x = 2

=> x = $\dfrac{4}{3}$ or x = 2

=> x = $1.\overline{33}$  or x = 2

Required answer -

=> Thus, the zeroes of the given polynomial will be $1.\overline{33}$ or 2 .

Verification -

⇔ We need to verify that the quadratic equation  3x²-10x+8​ = 0 is true for the values $x = \dfrac{4}{3} \:and\:x=2$ .

→ For  $x = \dfrac{4}{3}$

=> $3 \times\dfrac{4}{3}\times{\dfrac{4}{3} } -10 \times{\dfrac{4}{3} } + 8$

=> $4 \times\dfrac{4}{3} -{\dfrac{40}{3} } + 8$

=> $\dfrac{16}{3} -{\dfrac{40}{3} } + 8$

=> $\dfrac{16}{3} -{\dfrac{40}{3} } + \dfrac{8\times3}{1\times3}$

=> $\dfrac{16-40+24}{3}$

=> $\dfrac{40-40}{3}$

=> $\dfrac{0}{3}$

=> 0

=> R.H.S

∵ L.H.S = R.H.S = 0

∴ $\underline{\textbf{x = \dfrac{4}{3}\:satisfies\:the\:equation}}$$\underline{x = \dfrac{4}{3}\:\:satisfies\:\:the\:\:quadratic\:\:equation. }$

→ For x=2

=> 3 × 2 × 2 - 10 × 2 + 8

=> 12 - 20 + 8

=> 20 - 20

=> 0

=> R.H.S

∵ L.H.S = R.H.S = 0

∴ x =2 satisfies the quadratic equation.

=> Hence verified