# 5x squared  -3x +5 = 7 solve by factoring

sciencesolve | Certified Educator

You need to evaluate the quadratic equation, by factoring,  such that:

`5x^2 - 3x + 5 = 7`

You need to subtract 7 both sides, such that:

`5x^2 - 3x + 5 - 7 = 0`

`5x^2 - 3x - 2 = 0`

You need to solve for x the equation using factorization, hence, you may write the coefficient of x, such that:

`3 = 5 - 2`

Replacing 5 - 2 for 3, yields:

`5x^2 - (5 - 2)x - 2 = 0`

You need to remove the brackets, such that:

`5x^2 - 5x + 2x - 2 = 0`

Grouping the members yields:

`(5x^2 - 5x) + (2x - 2) = 0`

You need to factor out 5x in the first group and 2 in the second group, such that:

`5x(x - 1) + 2(x - 1) = 0`

You need to factor out x - 1, such that:

`(x - 1)(5x + 2) = 0` `=> {(x - 1 = 0),(5x + 2 = 0):} => {(x = 1),(x = -2/5):} `

Hence, solving the quadratic equation by factoring, yields `x = 1` and `x = -2/5.`

sid-sarfraz | Student

QUESTION:-

5x squared  -3x +5 = 7 solve by factoring

`5x^2-3x+5=7`

SOLUTION:-

This problem can be solved by two methods:-

• Factorization Method
• Quadratic Equation Formula

But according to the question we have to solve this problem by factorization;

Therefore;

`5x^2-3x-2=0`

`5x^2-5x+2x-2=0`

`5x(x-1)+2(x-1)=0`

Separating the two solution sets;

5x + 2 = 0                              ,                     x - 1 = 0

5x = -2                                      ,                     x = 1

x = -2/5

Hence the solution set is {-2/5, 1}

Hence Solved!

rachellopez | Student

When you factor, you have to first set your equation equal to zero. So, subtract the 7 from both sides

5x^2-3x-2=0

Now, find the factors that multiply to make 5x^2

(5x   )(x   )=0

and what factors multiply to make -2 and add to get -3x

(5x+2)(x-1)=0

When you factor, you will get the original equation, so you know it is correct. Lastly, set each factor equal to zero to Solve for x.

5x+2=0            x-1=0

5x=-2               x=1

x=-2/5

5x^2-3x+5-7=0

5x^2-3x-2=0

what multiplies to get -10 and adds to get -3?

-5*2

5X^2-5X+2x-2

put parenthesis

(5X^2-5X)+(2x-2)

take out 5x from the first one and 2 from the second

so you get

5x(X-1)+2(X-1)

then put out side ones together

(x-1) and (5x+2)

steamgirl | Student

Fist put into standard form and set equal to zero.

5x^2 -3x+5 =7

5x^2 - 3x -2 =0

Then factor using the box method (pic attached). Remember that standard form is ax^2 + bx + c = 0. So make a box with 4 squares, then put a in the inner box on the top left, and c in the lower right. Then work backwards to find a factored form that will multiply to make terms a and c, but will add to make b. You end up with (5x+2)(x-1). Which means x = 1, -2/5

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nisarg | Student

`5x^2-3x+5-7=0`

`5x^2-3x-2=0`

` `

(5x+2)(x-1)

atyourservice | Student

`5x^2-3x +5 = 7`  first combine like terms

`5x^2-3x +5-7= 7-7`

`5x^2-3x -2 = 0`

`a=5`  `b= -3`   `c=-2`

now multiply axxc

`5xx-2= -10 `  now find factors of -10 that add up to b  which woud be -5,2  plug those numbers into the equation

`5x^2-5x+2x-2 `   factor by grouping

`(5x^2-5x)(2x-2)`  find the greatest common factors

`5x(x-1) 2(x-1)`

(5x+2)(x-1)

Wiggin42 | Student

5x^2-3x+5=7

5x^2 - 3x - 2 = 0

(5x    a )(x      b ) = 0

You know that a and b have to be numbers so that

a x b = -2

5b + a = -3

You can fairly quickly run through possibilities. b must be negative since the sum has to be negative. If b must be negative, a must be positive so that their product is negative. The only possible values are 1 or 2 (ignoring negatives). With this we can reason that a = 2 and b = -1.

So, the original polynomial can be factored into: (5x + 2)(x -1). To check if you're correct, simply FOIL back out.

Solve for x:

(5x + 2) = 0        (x - 1) =0

x = -2/5

x = 1

piqachu | Student

5x^2-3x+5=7

or,5x^2-3x+5-7=0

or,5x^2-3x-2=0

or,5x^2-(5-2)x-2=0, here, you have to evaluate LCM of 10 (5×2)

or,5x^2-5x+2x-2=0

or,5x(x-1)+2(x-1)=0

or,(x-1)(5x+2)=0

here,x-1=0 or 5x+2=0

so,x=1 or x=-2/5

ans: x= 1 or -2/5