Question

find the zeroes of the polynomial 6x²-17x+12 and verify the relationship between zeroes and coefficient of polynomial​

Answer 1

Given :

• f(x) = 6² - 17x + 12

Then,

By using Middle term splitting method :

$:\implies \sf {f(x) = 6x^{2} - 17x + 12}$

$:\implies \sf{f(x) = 6x^{2} - (9 +8)x + 12}$

$:\implies \sf{f(x) = 6x^{2} - 9x - 8x + 12}$

$:\implies \sf{f(x) = 3x(2x - 3) - 4 (2x - 3)}$

$:\implies \sf{f(x) = (3x - 4)(2x - 3)}$

Now,

$:\implies \sf{ (3x - 4) = 0 }$

$:\implies \sf{ 3x = 4 }$

$:\implies \sf{ x =\dfrac{4}{3} }$

Or

$:\implies \sf{ (2x - 3) = 0 }$

$:\implies \sf{ 2x = 3}$

$:\implies \sf{ x =\dfrac{3}{2} }$

$:\implies \sf{ \alpha = \dfrac{4}{3} \: or \: \beta \: = \dfrac{3}{2}}$

α = 4/3 and β = 3/2 are the zeroes of the polynomial f(x).

• Given Polynomial = f(x) = 6² - 17x + 12

Let's compare it with the general form of quadratic equation :

$\bold{Quadratic \: Equation: \: ax^2+bx+c=0}$

• $\Rightarrow{ a = 6}$
• $\Rightarrow{ b = -17}$
• $\Rightarrow{ c = 12}$

Verifying relation between zeroes and coefficients

$\textsf{Sum of the Zeroes }$

$\longrightarrow\sf - {Coefficient \: of \: x/Coefficient \: of \: x^{2}}$

$:\implies \sf{ \alpha + \beta = \dfrac{17}{6}}$

$:\implies \sf{ \dfrac{4}{3} + \dfrac{3}{2} = \dfrac{17}{6}}$

$:\implies \sf{ \dfrac{(2 \times 4) + (3 \times 3)}{6} = \dfrac{17}{6}}$

$:\implies \sf{ \dfrac{8 + 9}{6} = \dfrac{17}{6}}$

$:\implies \bf{ \dfrac{17}{6} = \dfrac{17}{6}}$

⠀⠀$\underline{\bf{LHS = R.H.S}}$

_________________________

$\textsf{Product of the Zeroes }$

$\longrightarrow\sf{Constant \: Term/Coefficient \: of \: x^{2}}$

$:\implies \sf{ \alpha \times \beta = \dfrac{12}{6} }$

$:\implies \sf{ \alpha \times \beta = 2 }$

$:\implies \sf{ \dfrac{4}{3} \times {3}{2} = 2 }$

$:\implies \sf{ \dfrac{12}{6} = 2}$

$:\implies \bf{ 2 = 2 }$

⠀⠀$\underline{\bf{LHS = R.H.S}}$

Hence,

The relation between them also verified.

${\boxed{\bf{Verified}}}$

Answer 2

Answer:

$6 {x}^{2} - 17x + 12 \\ \\ = 6 {x}^{2} - 9x - 8x + 12 \\ \\ = 3x(2x - 3) - 4(2x - 3) \\ \\ = (2x - 3)(3x - 4) \\ \\ \\ \boxed{zeros = \frac{3}{2} \: \: and \: \: \frac{4}{3} }$