How to factorise ax + bx + ay + by.?
ax + bx + ay + by = (ax + bx) + (ay + by) = x(a + b) + y(a + b) = (a + b)(x + y)
Factorise ax + ay + bx + by. help please :P?
You can factor by grouping here: a(x + y) + b(x + y) = (a + b)(x + y) http://www.basic-mathematics.com/factori...
Factoring help, ax+bx+ay+by+az+bz?
Group (ax + bx) + (ay + by) + (az + bz) Common factors x(a + b) + y(a + b) + z( a + b) Note a+ B is a common factor (x + y + z)(a + b) is the factored result.
How do you factor ax+ay+3bx+3by?
Factor like terms.. (ax + ay) + (3bx + 3by) = a(x + y) + 3b(x + y) Finally, group the common factors of x + y and collect the terms to get: (x + y )(a + 3b) I hope this helps!
How do you factorise ax+by+bx+ay (WITH WORKING OUT PLEASE)?
x(a + b) + y(a + b) = (a + b)(x + y)
Factor ax²+bx+c...Please help?
2x²+7x+3 3*2 = 6 Find two numbers whose product is 6 and sum is 7. The numbers are 1 and 6 2x²+6x+x+3 = 2x(x+3)+1(x+3) = (2x+1)(x+3) 8x²+18x+9 8*9 = 72 Find two numbers whose product is 72 and sum is 12. The numbers are 12 and 6 8x²+12x+6x+9 = 4x(2x+3)+3(2x+3) = (4x+3)(2x+3) 6x²-9x+5 6*5 = 30 Find two numbers whose product is 30 and sum is -9. There are no such numbers. This cannot be factorised unless you allow for complex numbers. 3k²+32x-11 3*(-11) = -33 Find two numbers whose product is -33 and sum is 32. The numbers are 33 and -1 3k²-k+33k-11 = k(3k-1)+11(3k-1) = (k+11)(3k-1) 18n²+9n-14 18*(-14) = -252 Find two numbers whose product is -252 and sum is 9. The numbers are 21 and -12 18n²-12n+21n-14 = 6n(3n-2)+7(3n-2) = (6n+7)(3n-2) 12v²-25v-7 12*(-7) = -84 Find two numbers whose product is -84 and sum is -25. The numbers are -28 and 3 12v²-28v+3v-7 = 4v(3v-7)+1(3v-7) = (4v+1)(3v-7) = (6n+7)(3n-2)
Why are the factors of a^2-b^2 not (a x a)-(b x b)?
a^2-b^2=(a*a)-(b*b),It is also a factor.The other factors are (a+b) and (a-b)
How do I factor a cubic that looks like ax^3 + bx^2 + cx =0 but that does not have the d?
[math]ax^3 + bx^2 + cx = 0 [/math]…is a good and fair one!Now, given that this is freshman level mathematics, I assume you’ve heard of the null factor law, i.e:[math](x-d)(x-e) = 0[/math]From this law, we can assume that both the first product and the second product are equal to zero, or[math]x=d,e[/math]Then, we can keep this law in the back of our heads and chuck out an x somewhere in that above cubic, so that it becomes:[math]x(ax^2 + bx + c) = 0[/math]So now we have two equations:[math]x=0, x^2 + bx/a + c/a[/math]For the quadratic, we then use that null factor law to find any solutions that add into the first degree coefficient, and multiply into the constant.Or, we use the quadratic formula:[math]x= -b/2a + sqroot(b^2-4ac)/2a, -b/2a - sqroot(b^2-4ac)[/math]There we get the remaining solutions, real or complex.Note, complex solutions in the form:[math]x=a+bi[/math] can be expressed in polar form, or:[math]x=r*cis(theta). [/math]These are much easier to work with.
How can quadratic equations be solved by factoring?
There are a few std quick-trick types:⑴ PRODUCT & SUM (ALWAYS think of PRODUCT FIRST)x²±(A+B)x÷AB=0ex. x²±12x+35=0x²±(7+5)+(7×5)=0→→(x±5)(x±7)=0 →→x=±5,x=±7⑵ PRODUCT & DIFFERENCEx²±(A-B)x-ABx²±2x—35=0x²±(7–5)x—(7×5)=0(x+5)(x—7)=0 or (x—5)(x+7)=0⑶FRONT + BACK =MIDDLEAx²±(A+B)x+B=0(Ax±B)(x±1)=07x²±12x+5=0(7x±5)(x±1)=0 ......x=±1 orx=±5/7⑷ ( FRONT×BACK)±1=MIDDLE7x²±36x +5=0 or 7x²±34x—5=0(7x±1)(x±5)=0,(7x—1)(x+5)=0GOOD LUCK!⑷FRONT
How do I factorize a quartic equation? And how do I factorize any polynomial that's not in its standard form (ax + bx + cx���)? By that I mean if any of the constant integers are missing, like b, c, d or e= 0.
It is very difficult to.When you have a polynomial in the form ax^4+bx^3+cx^2+dx+e, it could possibly have the roots in the form of p/q, where p is a factor of e and q is a factor of a. (With megative values included.)To find the correct p and q you plug the number in and see if the expression evaluates to 0. If it does, then it has a factor (qx-p). Then you do synthetic division and turn it into a cubic.Usually there are no rational x that it evaluates to 0. If so, there is no p or q that satisfies. You only know when you tried every single value of p and q.