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# Can You help me with Solving Quadratic Equations by Factoring?

Topic: Problem solving by quadratic equations
May 26, 2019 / By Ava
Question: I'm learning solving quadratic equations by factoring can you help me with sum of these problems please & thanks you :) 1. x²+4x-12 = 0 2. x²-4x=21 3. n²-4= -3n 4. 3x²+13x+4=0 5. 3p²+8p+5=0 6. 3r²+10r+9=6 7. 2n²+17n+11=3 Thanks

## Best Answers: Can You help me with Solving Quadratic Equations by Factoring?

Abby | 7 days ago
x^2+4x-12=0 factors of twelve that will give you a middle term of 4x are 6 and 2 so factors are (x+6)(x-2) pay attention to signs if you need to solve for x, then set bothe parenthesis to =0 and x=-6, 2 and that's how you should do the rest of the problems ))
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We found more questions related to the topic: Problem solving by quadratic equations

You should have the following patterns so drilled into your head by countless homework exercises that you recognize them whether the numbers are constant or variables. a^-b^2 = (a-b)(a+b) (a+b)^2 = a^2 + 2ab + b^2 (a-b)^2 = a^2 -2ab + b^2 In high school this is the same as 3 time 7 equals 21 in third grade. If you plan to learn math without doing any F**** homework, now is the time to use the technique you planned to substitue.
1. (x+2)(x-6)=0 x+2=0 or x-6=0 (below) (subtract 2 from both sides) x = -2 or x-6=0 (add six to both sides) x = 6 so, x= -2 or 6 2. (x-2)(x-4)=0 x-2=0 or x-4=0 (same procedure as above) x=2 or x= 4

Steph
1. x²+4x-12 = 0 (x-2)(x+6)=0 2. x²-4x=21 0=x^2-4x-21 then, (x+3)(x-7)=0 3. n²-4= -3n 0=n^2+3n-4 then, (n-1)(n+4) 4. 3x²+13x+4=0 (3x+1)(x+4)=0 5. 3p²+8p+5=0 (3p+5)(p+1)=0 6. 3r²+10r+9=6 = 3r^2+10r+3 so, 0=(3r+1)(r+3) 7. 2n²+17n+11=3 =2n^2+17n+8 so, 0=(2n+1)(n+8)
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