Question

Find the zero of the quadratic polynomial x2 + 7x +10 and verify the relationship between the zeroes and the coefficients

Answer 1

Hey mate, here is your answer:-

= x2+7x+10

= x2+(5+2)x+10

= x2+5x+2x+10

= x(x+5) +2(x+5)

= (x+2)(x+5)

The zeros of the polynomial are:- -2 and -5...

VERIFICATION:-

Sum of zeros= -b/a
-2+(-5)= -7/1
-2-5= -7
-7= -7

Product of zeros= c/a
-2×-5= 10/1
10= 10

HENCE VERIFIED....

HOPE it helps!!!!
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Answer 2

Answer:

-3,-4

Step-by-step explanation:

x2 + 7x +12

factorise the equation:

x2 + 4x + 3x + 12

x(x+4) + 3(x+4)

= (x+3)(x+4)

= x= -3 and x= -4

the two zeroes are -3 and -4

verification of relationship

As you know:

ax2 + bx + c

alpha= -3 and beta= -4

alpha + beta = -b/a

so, -3-4= -b/a

-7= -b/a

minus and minus be cut off

7/1= b/a

b= 7

a= 1

now, alpha x beta = c/a

-3 x -4 = c/a

12/1= c/a

c= 12

a= 1

therefore, a= 1, b= 7, c= 12

so as per formula: ax2 + bx + c

x2 + 7x + 12

hope you understood this!!!

all the best for your future!!!