Find the zeroes of the polynomial 5t2 + 12t + 7, and verify the relati

1.	Find the zeroes of the polynomial 5t2 + 12t + 7, and verify the relation between the coefficients and the zeroes of the polynomial.
Answer» 5t² + 12t + 7 Splitting the middle term, we get, 5t² +5t + 7t + 7 Taking the common factors out, we get, 5t (t+1) +7(t+1) On grouping, we get, (t+1)(5t+7) So, the zeroes are, t+1=0 ⇒ y= -1 5t+7=0 ⇒ 5t=-7⇒t=-7/5 Therefore, zeroes are (-7/5) and -1 Verification: Sum of the zeroes = – (coefficient of x) ÷ coefficient of x² α + β = – b/a (- 1) + (- 7/5) = – (12)/5 = – 12/5 = – 12/5 Product of the zeroes = constant term ÷ coefficient of x² α β = c/a (- 1)(- 7/5) = – 7/5 – 7/5 = – 7/5

Find the zeroes of the polynomial 5t2 + 12t + 7, and verify the relation between the coefficients and the zeroes of the polynomial.

Answer»

5t² + 12t + 7

Splitting the middle term, we get,

5t² +5t + 7t + 7

Taking the common factors out, we get,

5t (t+1) +7(t+1)

On grouping, we get,

(t+1)(5t+7)

So, the zeroes are,

t+1=0 ⇒ y= -1

5t+7=0 ⇒ 5t=-7⇒t=-7/5

Therefore, zeroes are (-7/5) and -1

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

(- 1) + (- 7/5) = – (12)/5

= – 12/5 = – 12/5

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 1)(- 7/5) = – 7/5

– 7/5 = – 7/5