How can I solve by factorising 3x^2 - 7x - 6?
We can’t “solve” because it is not an equation, but we can factor it.We are looking to factor [math]3x^2-7x-6[/math]I like to use the ac method.First, you multiply the [math]x^2[/math] coefficient and the constant term. That gives us [math](3)(-6)=-18[/math] in this case.Now, you need to find 2 numbers that multiply to that same value, but they add to the x coefficient of -7.18 doesn’t have many factors, so it isn’t too hard to determine. The only pairs we would have would be 1 and 18, 2 and 9, and 3 and 6. Since they need to multiply to make a negative value, one must be negative and one must be positive. Since they need to add to a negative value, the negative one must be the larger number. That narrows us down to -18 and 1, or -9 and 2, or -6 and 3. Checking those, the middle one is the correct one, -9 and 2. Those numbers check out because they multiply to -18 and add to -7.Now that we have those 2 numbers, we use those to split up the middle term in the original expression.[math]3x^2-7x-6[/math][math]3x^2-9x+2x-6[/math]Now we can apply factoring by grouping. We put the first 2 terms in parenthesis, then we add the last 2 terms in parenthesis.[math](3x^2-9x)+(2x-6)[/math]We factor each group individually. The greatest common factor in the first group is 3x. The greatest common factor in the second group is 2.[math]3x(x-3)+2(x-3)[/math]Now we factor out the shared binomial factor, in other words we factor out the [math](x-3)[/math].[math](x-3)(3x+2)[/math]The expression you listed was not an equation, so I cannot “solve” it, but this is how you can factor it somewhat easily.
Factorize x³-6x² +3x+ 10?
Apply trial and error to the equation by substituting x as any random integer to the equation [math] x^3 - 6x^2 + 3x + 10 = 0 [/math]. If the equation gives a result of 0, then (x - root) will be one factor of the equation. In that case, consider x = 2, so [math] 2^3 - 6(2)^2 + 3 (2) + 10 = 0 [/math] [math] 8 - 24 + 6 + 10 = 0 [/math] Hence, (x - 2) is one of the roots of the equation. Now, apply long division, and accordingly we will obtain another factor as [math] x^2 - 4x - 5 = (x - 5) (x + 1) [/math]. Hence, [math] x^3 - 6x^2 + 3x + 10 = (x + 1) (x - 2) (x - 5) [/math].
FACTOR: x² - 2x -3 / 3x² + x -2?
x^2 - 2x - 3 =(x + 1)(x - 3) And: 3x^2 + x - 2 = (3x - 2)(x + 1) The factor (x + 1) cancels out, leaving (x - 3)/(3x - 2).
How do you factor [math]x^3-6x^2+11x-6=0[/math]?
Given (x^3)-6(x^2)+11x-6=0x^3-3x^2-3x^2+11x-6=0x^2(x-3)-(3x^2-11x+6)=0x^2(x-3)-(3x^2-9x-2x+6)=0x^2(x-3)-(3x(x-3)-2(x-3))=0x^2(x-3)-[(x-3)(3x-2)]=0(x-3)(x^2-3x+2)=0(x-3)(x^2-2x-x+2)=0(x-3)(x(x-2)-1(x-2))=0(x-3)(x-2)(x-1)=0Online Math Tutor
How do you factor [math]x^4 -x^3 -7x^2 +x +6[/math]?
Always try x = 1.Just to test, we see1^4 - 1^3 - 7*1^2 + 1 + 6 = 0.So 1 is a root. We can factor out (x-1) to getx^4 - x^3 - 7x^2 + x + 6 = (x-1)(x^3 - 7x - 6).Always try x = -1.(-1)^3 -7(-1) + 6 = 0.So -1 is a root. Factor that out to get(x-1)(x^3 - 7x - 6) = (x-1)(x+1)(x^2 - x - 6).The last term is a quadratic that factors, sox^4 - x^3 - 7x^2 + x + 6 =(x-1)(x+1)(x-3)(x+2)
3x^2+7x-6=0 this factoring problem is confusing me. PLEASE HELP!?
3x² + 7x - 6 = 0 x² + 7/6x = 2 + (7/6)² x² + 7/6x = 72/36 + 49/36 (x + 7/6)² = 121/36 x + 7/6 = ± 11/6 = x + 7/6 - 11/6, = x - 4/6, = x - 2/3, = 3x - 2 = x + 7/6 + 11/6, = x + 18/6, = x + 3 Answer: (3x - 2)(x + 3) are the factors. Proof (F.O.I.L.): = (3x - 2)(x + 3) = 3x(x) + 3x(3) - 2(x) - 2(3) = 3x² + 9x - 2x - 6 = 3x² + 7x - 6
Factor this please?
First factor out x: 2x³-3x²-12x = x·(2x²-3x-12) then factor (2x²-3x-12) An easy way to factor "naughty" quadratics is to factor out the leading coefficients and then use the quadratic formula: !!! (but do not forget to first factor the leading coefficient [the coeff of x²] out, or you are going to lose this factor in your final answer) !!! (2x²-3x-12) = 2(x² -3/2x - 6) x = [3/2 ± √(9/4 + 36 )] / 2 x = [3/2 ± √(153/4)] / 2 x = [3/2 ± (√153)/2)] / 2 x = [3 ± (√153)] / 4 x = [3 ± (3√17)] / 4 x = 3/4(1± √17) So (2x²-3x-12) = 2(x² -3/2x - 6) = 2(x - 3/4(1+ √17))(x -3/4(1- √17)) And 2x³-3x²-12x = x·(2x²-3x-12) = 2·x·(x - 3/4(1+ √17))·(x -3/4(1- √17)) Not a very elegant result :-/ Either the question really is "funny" or there is a typo :-)
If x+a is a factor of x^3+ax^2-2x+a+4 then a equals?
That is a very easy question.Theorem To solve this question, we would need to apply the Factor Theorem.If you don't know the Factor Theorem:Factor TheoremIt states that if f(x) is a polynomial and g(x) is the factor of f(x), where the degree of g(x) must be smaller than f(x), then the value/values of x when g(x) is equated to zero is/are the zeroes of the polynomial.As f(x)=g(x)*q(x) + 0 ,[q(x) is the other factor/factors]Now, if for any value of x g(x)=0Then, f(x) will also be equal to zero.Thus we get a theorem to solve your question.If you know the Factor Theorem, The Real Answer Starts HERE:Now that you know the key theorem this question becomes pretty easy.Here, f(x)= x^3 + ax^2- 2x + a + 4And, g(x)= x + aWe have, x + a = 0=> x = -aNow, put the value thus found into f(x) functionf(-a) = (-a)^3 + a(-a)^2 - 2(-a) + a + 4 =0=> -a^3 + a^3 + 2a + a = -4=> 3a = -4=> a = -4/3
If (x-2) and (x-3) are factors of 2x^2-mx+n, then what are the values of m and n?
As (x-2) and (x-3) are factors of given function f(x) = 2x^2 - mx + n(x-2) (x-3) = f(x)x*x - 2x - 3x + 6 = f(x)x^2 - 5x+ 6 = 2x^2 - mx + nSo, m = 5*2 = 10Andn = 6*2 = 12.So values of m and n are 10 and 12 respectively.