Question

find the zero of polynomial and verify
a) 3x²-17x+24
b) 6x²-17x+7
c) x²-12x+32

Answer 1

Answer:-

a) 3x² - 17x + 24 = 0

3x² - 9x - 8x + 24 = 0

3x (x - 3) -8(x - 3) = 0

(3x - 8) (x - 3) = 0

3x - 8 = 0

3x = 8

x = 8/3.

x - 3 = 0

x = 3.

b) 6x² - 17x + 7 = 0

6x² - 3x - 14x + 7 = 0

3x (2x - 1) -7(2x - 1) = 0

(3x - 7)(2x - 1) = 0

3x - 7 = 0

3x = 7

x = 7/3.

2x - 1 = 0

2x = 1

x = 1/2.

c) x² - 12x + 32 = 0

x² - 4x - 8x + 32 = 0

x(x - 4) - 8(x - 4) = 0

(x - 8)(x - 4) = 0

x - 8 = 0

x = 8

x - 4 = 0

x = 4.

Verification:

a) Sum of zeroes = -b /a

8/3+3 = -(-17/3)

(8+9)/3 = 17/3

17/3 = 17/3.

Product of zeroes = c/a

(8/3)(3) = 24/3

24/3 = 24/3

b) Sum = -b/a

(7/3)+(1/2) = 17/6

2(7)+3(1)/6 = 17/6

17/6 = 17/6.

Product = c/a

(7/3)(1/2) = 7/6

7/6 = 7/6

c) Sum = -(-12)/1

4+8 = 12

12 = 12.

Product = 32/1

(8)(4) = 32

32 = 32.