Can Anyone Help Me With My Summer honors Geometry packet?

Question

Hi i was wondering if anyone could possibly help me with Factoring Quadratic Equations

i've done most of them but i have gotten stuck on some could you help me i don't need the answer i just need someone to show me how you solved this problem

7a^2+22a+3=0

and also a^2+a-30=0

yes i know these are very easy i just totally forgot how to do this over my summer Thanks for your help

Bianca

Answer 1

7a^2 + 22a + 3 = 0

The coefficient of the first degree term is 22. The coefficent of the other two terms are 7 and 3. Find two numbers with a sum of 22 and a product of (7)(3)=21

The numbers are 21 and 1.

So we replace the 22a with those two numbers.

7a^2 + 21a + 1a + 3 = 0

Bracket the first two and last two into two binomials.

(7a^2 + 21a) + (1a + 3) = 0

Factor out 7a from the first binomial

7a(a + 3) + (a + 3) = 0

Using distributive property

(7a + 1) * (a + 3) = 0

This tells us that a = (-1/7) or a = (-3)

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a^2 + a - 30 = 0

The coefficient of the first degree term is +1. The coefficent of the other two terms are +1 and -30. Find two numbers with a sum of 1 and a product of (1)(-30)=(-30)

The numbers are +6 and -5.

So we replace the +a with those two numbers.

a^2 + 6a - 5a - 30 = 0

Bracket  the first two and last two into two binomials. Notice the sign change in the second binomial because of the brackets and the subtraction between the binomials.

(a^2 + 6a) - (5a + 30) = 0

Factor out +a from the first binomial

a(a + 6) - (5a + 30) = 0

Factor out 5 from the second binomial.

a(a + 6) - 5(a + 6) = 0

Using distributive property

(a - 5) * (a + 6) = 0

This tells us that a = 5 or a = (-6)