Algebra Factorising?

hey!

Although these look really difficult, you need to tackle them one step at a time.

1) First look at the numbers in front, do they have a factor that goes into them?
15,20,35? yes, 5 is the biggest no that goes in.
so on the outside before brackets you put 5, since you will multiply with whats in the brakets.
2) Now, in all of the equations with letters appear?
they all have x and y only one equation has z so it doesnt go outside since no other equations have it.
BUT hold on, to save you time you can see that although these appear you can (square (^) ) them so its easier in the equation.which one can you do that to x (on the outisde)

here it the answer-

5x^2y(3x^2+4y^2z-7xy)

get it?? Try multiplying it out, it will be what you started with.

Using the method i just said try doing the other one

The answer is=
7xy^2(2xy+3-5x^2y^2)

You need the most 'common' thingys outside the brackets to make your life easier when multiplying with whats inside.
so if you include squares on the outside, which can appear inside aswel, then you can still manage to multiply.

REMEMBER-with indices when you MULTIPLY you ADD the powers so if on the outside it says,
x^3(x^3) You add them so its x^6


I hope it makes sense!!
:)

those are the first parts of quadratic equations, and the following is the thanks to do them. i have examine the different solutions; they're all authentic, yet in case you quite warfare with maths it is how i develop into taught and that i imagine that's the way in which there is. evaluate x² - 11x + 30. placed 2 pairs of brackets less than it, with an x in, like so: (x )(x ) Then that's factorising. For a number of those equation you want 2 numbers that multiply to make the top type of the equation, so accordingly multiply to make 30. Then that's all about the indications. the first signal is the signal of the finest of both numbers, so the following that's a -. the 2d signal tells you 2 issues; first, it tells you the signal of the smaller variety, if that's + then the signal is an same because the signal for the better variety, if that's - then both indicators are diverse. So this leads us to (x- )(x- ). Secondly it tells you the numbers. If the signal is - then both numbers must have a huge difference of the middle type of the equation, if the signal is + then both numbers could upload as a lot as make the middle type of the equation. So for x² - 11x + 30 we favor 2 numbers that multiply to make 30 and upload as a lot as 11. It must be 6 and 5. Then basically placed them contained in the brackets (remembering that the first signal of the equation is the signal of the better variety) so we've (x-6)(x-5) because the answer. attempt the first one back following those regulations, you're answer could be (x-10)(x+2). you're able to be in a position to entice close those with the intention to progression onto calculus as they're a key portion of it. desire that enables.

In the previous answers, the 7xy^2(2xy + 3 - 5x^2y^2) factorises even further. You can see this by noting that the bracketed term (2xy + 3 - 5x^2y^2) is itself a quadratic in xy. If z=xy, it is

2z + 3 - 5z^2 = 3 + 2z - 5z^2 = (1 - z)(3 + 5z)

Thus the answer is 7xy^2(1 - xy)(3 + 5xy).

1)
15x^4y + 20x^2y^3z - 35x^3yz^2
= 5(3x^4y + 4x^2y^3z - 7x^3yz^2)
= 5x^2y(3x^2 + 4y^2z - 7xz^2)

2)
14x^2y^3 + 21xy^2 - 35x^3y^4
= 7(2x^2y^3 + 3xy^2 - 5x^3y^4)
= 7xy^2(2xy + 3 - 5x^2y^2)

When you know square means even powers, first fish out all powers which are odd.
In your problem 5x^3yz(3x+4y^2-7z) is NOT a difference of 2 squares. Neither is the second.

I can't help but I just wanted to say how adroit you must be!

try try try