Question

5x² - 4-8x middle term split​

Answer 1

Answer:

Hey!

Given polynomial :- 5x² - 4 - 8x

Any Quadratic polynomial should be in the form of ax² + bx + c

So, the polynomial is 5x² - 8x - 4

Let's factorise it by middle term splitting method :)

5x² - 8x - 4

5x² - 10x + 2x - 4

5x ( x - 2 ) + 2 ( x - 2 )

( 5x + 2 ) ( x - 2 )

• ( 5x + 2 ) = 0

x = -2/5

• ( x - 2 ) = 0

x = 2

So, the zeros are -2/5 and 2 !!

° To verify :-

• Sum of Zeros =

= \frac{ - coeff. \: \: of \: x}{coeff. \: of \: {x}^{2} }=

coeff.ofx

2

−coeff.ofx

Taking LHS

° Sum of Zeros = -2/5 + 2

\begin{gathered} = \frac{ - 2 + 10}{5} \\ \\ = \frac{8}{5} \end{gathered}

=

5

−2+10

=

5

8

Now ,taking RHS

\frac{ - coeff. \: \: of \: x}{coeff. \: of {x}^{2} }

coeff.ofx

2

−coeff.ofx

= \frac{ - ( - 8)}{5} = \frac{8}{5}=

5

−(−8)

=

5

8

Hence , LHS = RHS !!

• Product of Zeros =

\frac{constant \: term \: }{coeff. \: \: of \: {x}^{2} }

coeff.ofx

2

constantterm

Taking LHS

° Product of Zeros = -2/5 × 2 = -4/5

Now, taking RHS

\frac{constant \: term}{coeff. \: \: of \: {x}^{2} }

coeff.ofx

2

constantterm

= \frac{ - 4}{5}=

5

−4

Hence in both the case LHS = RHS

so, it's Verified !!