Question

factorize by splitting the middle terms a^2+6a+8

Answer 1

Answer:

$\huge{\ornage{\boxed{\boxed{\pink{\underline{\green{\mathscr{Answer}}} }} }}}$

a² + 6a + 8

a² + 2a + 4a + 8

a(a + 2) + 4(a + 2)

(a + 4)(a + 2)

Step-by-step explanation:

Answer 2

Answer:

$\bigstar{\bold{Zeros\:are\:-4\:and\:-2}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

p(x) = a² + 6a + 8

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

Zeros of the polyomial by splitting the middle term

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ p(x) = a² + 6a + 8

→ By splitting the middle term,

  a² + 4a + 2a + 8 = 0

→ Taking the common factors out,

  a ( a + 4 ) + 2 (a + 4 ) = 0

→ Taking a+4 as common,

  ( a + 4 ) ( a + 2 ) = 0

→ Either

  a + 4 = 0

  a = -4

→ Or

  a + 2 = 0

  a = -2

→ Hence the zeros are -4 and -2

$\boxed{\bold{Zeros\:are\:-4\:and\:-2}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The zeros of a quadratic polynomial can be found out by:

• Factorization method
• Splitting the middle term
• Completing the square method