Question

answer fast guys plz I knew so easy question but I forgot some steps how to solve

x⁴+2x³-2x²+2x-3 is divisible by
x²+2x-3.​

Answer 1

Answer:

Step-by-step explanation:x² + 2x - 3

=> x² + 3x - x - 3

=> x(x + 3) - (x + 3)

=> (x + 3)(x - 1)

Now, by remainder theorem, remainder = 0

Taking, x + 3 as the factor

So, x = -3

x⁴ + 2x³ - 2x² + 2x - 3 = 0

(-3)⁴ + 2(3)³ - 2(-3)² + 2(-3) - 3 = 0

81 - 54 - 18 - 6 - 3 = 0

81 - 54 - 27 = 0

81 - 81 = 0

0 = 0

Hence, (x + 3) is the factor of given equation,

Now, checking for (x - 1) as a factor,

So, x = 1

x⁴ + 2x³ - 2x² + 2x - 3 = 0

(1)⁴ + 2(1)³ - 2(1)² + 2(1) - 3 = 0

1 + 2 - 2 + 2 - 3 = 0

3 - 3 = 0

0 = 0

Then, x - 1 is Also a factor,

As both (x - 1) and (x + 3) are factors, we can say that the given equation is divisible by x² +2x - 3

Hence, proved.

Step-by-step explanation:

Answer 2

Dividing (x⁴+2x³-2x²+2x-3) by (x²+2x-3)

SOLUTION:

$\boxed{\begin{array}{ c |c cccc}\cline{2-3}&(x^{2}&+1)\\ \cline{1-6} x^{2}+2x-3 & x^{4} & +2x^{3} &-2x^{2}&+2x&-3 \\ &(-)x^{4}&(-)+2x^{3}&(+)-3x^{2}\\ \cline{2-6} &&&1x^{2}&+2x&-3 \\ &&&(-)1x^{2}&(-)+2x&(+)-3\\ \cline{2-6} &&&&0&\\ \end{array}}$

Now,

Now finding other factors,

x² + 2x - 3

x² + 3x - 1x - 3

x(x+3) -1(x+3)

(x - 1)(x+3)

Hence,

Factors are (x - 1), (x+3), and zeroes are (1) and (-3)

Verification:

P(x) = x⁴ + 2x³ - 2x² + 2x - 3

P(1) = (1)⁴ + 2(1)³ - 2(1)²+ 2(1) - 3

P(1) = 1 + 2 - 2 + 2 - 3

P(1) = 5 - 5

P(1) = 0

So, x = 1 is a zero

Now,

P(x) = x⁴ + 2x³ - 2x² + 2x - 3

P(-3) = (-3)⁴ + 2(-3)³ + 2(-3)²+ 2(-3) - 3

P(-3) = 81 + 2(-27) + 2(-9)+ 2(-3) - 3

P(-3) = 81 - 54 - 18 - 6 - 3

P(-3) = 81 - 81

P(-3) = 0

So, x = (-3) is a zero