Answer: Read Sol.
Step-by-step explanation: To solve these, call each given zero = z. The function is when given a zero that is an integer (x-z1)(x-z2)(x-z3)... so on until all zeroes are used. This works for fractions also. But when you have irrationals, you will have to make sure each irrational solutions conjugate is there. So if a given zero is 2-sqrt3, then always 2+sqrt3 will also be a 0. Likewise, if given -4-sqrt7, then -4+sqrt7 will also always be a zero. Use this logic to solve them very quickly! I hope this helps!
Factorise 4b² + 16b ⇒ 4b(b+4)
expand (a-1)(a-2) ⇒ a(a-2) - 1(a-2) ⇒ a² - 2a - 1a + 2 ⇒a² - 3a + 2
factorise x² + 8x + 7 ⇒(x + 1) (x + 7)
evaluate y⁶ / y² ⇒ y * y * y * y * y * y / y * y = y⁴
if x = -1 and y = 5, find z when z = x² + 2y² ;
z = -1² + 2(5²) ⇒ 1 + 2(25) ⇒1 + 50 = 51
Make x the subject: y = 4x - 3
y = 4x - 3
<u>+3 +3</u>
3 + y = 4x
<u>÷4 ÷4 </u>
(3+y)/4 = x
The perimeter is 30 for this answer I just did this yesterday. What you do is look at the top of the triangle the other side of where 6 is at is congruent and you do the same for 4 and 5
