Question

if -5 and 7 are zeroes of x⁴ - 6x³-26x²+138x -35 find the other number ?
PLEASE HELP ME..​

Answer 1

Step-by-step explanation:

(x+5)(x-7) = 0

x² - 2x - 35 = 0

_________________

x²-2x-35√ x⁴-6x³-26x²+138x-35 \_x²-4x+1

-x⁴+2x³+35x²

--------------------------------

-4x³+9x²+138x-35

+4x³-8x²-140x

---------------------------

x² -2x - 35

-x² + 2x + 35

-------------------

0 0

-------------------

x² - 4x + 1 = 0

x = 4 + √16-4 / 2 = 4 + 2√3 / 2

x = 2+√3 , 2-√3

Answer 2

Given:-

P(x)=x⁴-6x³-26x²+138x-35

Zeros=-5&7

To Find:-

The other Zeros?

Proof:-

→(x+5)=0 [°•° -5 is one of the zeros]

→(x-7)=0 [°•° 7 is one of the zeros]

★Multiply the Zeros★

→(x+5)(x-7)=0

→x²-7x+5x-35=0

→x²-2x-35=0

★Divide the Equation by p(x)★

→x²-4x+1=0–-[Is the Equation we get]-–

★Using Discriminant Method★

→d=b²-4ac

a=1,b=-4,c=1

→d=(-4)²×-4×1×1

→d=16-4

→d=12

→√d=√12

→√d=±2√3

$\rule{200}{1}$

•Taking x(+)=$\frac{-b+ \sqrt{d}}{2a}$

→x=$\frac{-(-4)+2 \sqrt{3}}{2×1}$

→x=$\frac{4+2 \sqrt{3}}{2}$

→x$\frac{\cancel{2}(2+ \sqrt{3})}{\cancel{2}}$

→x=$\frac{2+ \sqrt{3}}{1}$

→x=2+√3

$\rule{200}{1}$

•Taking x(-)=$\frac{-b- \sqrt{d}}{2a}$

→x=$\frac{-(-4)-2 \sqrt{3}}{2×1}$

→x=$\frac{4-2 \sqrt{3}}{2}$

→x$\frac{\cancel{2}(2- \sqrt{3})}{\cancel{2}}$

→x=$\frac{2- \sqrt{3}}{1}$

→x=2-√3

$\rule{200}{7}$