7x^2-2x+1-(9x^2+2x-1)

Mathematics

2 answers:

likoan [24]3 years ago

6 0

Answer:

$-2x^2-4x+2$

Step-by-step explanation:

$(7x^2-2x+1)-(9x^2+2x-1)=\\(7x^2-9x^2)+(-2x-2x)+(1+1)=\\-2x^2-4x+2$

Hope this helps!

zalisa [80]3 years ago

6 0

(7x2-2x-1)+(9x2-4x+5)-(4x2+6x-7)

Final result :

12x2 - 12x + 11
Step by step solution :

Step 1 :

Equation at the end of step 1 :

((((7•(x2))-2x)-1)+(((9•(x2))-4x)+5))-((22x2+6x)-7)
Step 2 :

Equation at the end of step 2 :

((((7•(x2))-2x)-1)+((32x2-4x)+5))-(4x2+6x-7)
Step 3 :

Equation at the end of step 3 :

(((7x2-2x)-1)+(9x2-4x+5))-(4x2+6x-7)
Step 4 :

Trying to factor by splitting the middle term

4.1 Factoring 12x2-12x+11

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If x+1/x= 3, then prove that m^5+1/m^5= 123

alina1380 [7]

9514 1404 393

Explanation:

We can start with the relations ...

$\displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)$

$\displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}$