Question

Find the zeroes of the quadratic polynomial 4s2−4s+1" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: 400; line-height: normal; font-size: 24px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: 0px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; color: rgb(134, 71, 102); font-family: HelveticaNeue; font-variant-ligatures: normal; font-variant-caps: normal; orphans: 2; widows: 2; -webkit-text-stroke-width: 0px; background-color: rgb(255, 241, 234); text-decoration-style: initial; text-decoration-color: initial; position: relative;">4s2−4s+1

Answer 1

Question :

Find the zeroes of the quadratic polynomial 4s²−4s+1

Answer:

1/2 and 1/2 are the zeroes of the given polynomial.

Given :

quadratic polynomial, 4s² - 4s + 1

Solution :

Given quadratic polynomial,

4s² - 4s + 1

To solve it using factorization method,

     we must know the sum - product pattern

• 4s² - 4s + 1

=> It is of the form ax² + bx + c

Find the product of quadratic term [ax²] and constant term [c]

= 4s² × 1

= 4s²

Now, find the factors of "4s²" in pairs

$=> s \times 4s \\\\ => -s \times -4s\\\\ => 2s \times 2s \\\\ =>-2s \times -2s$

from the above, find the pair that adds to get linear term [bx]

-2s - 2s = -4s

Now split -4s as -2s and -2s,

  4s² - 4s + 1 = 0

 4s² - 2s - 2s + 1 = 0

Find the common factor,

2s(2s - 1) - 1(2s - 1) = 0

(2s - 1)(2s - 1) = 0

=> (2s - 1) = 0 ; s = 1/2

$\boxed{\bf \frac{1}{2} \ and \ \frac{1}{2} \ are \ the \ zeroes \ of \ the \ polynomial}$

Answer 2

Step-by-step explanation:

Hope my answer is helpful to u