Question

1. write in factored form

$2v^2 + 11v + 5$
factored form:_________

1. $7a^2 + 53a = -28$

solutions :________
factors:________

Explanation:

Step1: Split middle term for $2v^2+11v+5$

Find two numbers: $10$ and $1$, since $10\times1=2\times5=10$, $10+1=11$.
$2v^2 + 10v + v + 5$

Step2: Factor by grouping

$2v(v+5) + 1(v+5)$

Step3: Extract common binomial factor

$(2v+1)(v+5)$
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Step4: Rewrite equation 4 in standard form

Add $28$ to both sides: $7a^2 + 53a + 28 = 0$

Step5: Split middle term for quadratic

Find two numbers: $49$ and $4$, since $49\times4=7\times28=196$, $49+4=53$.
$7a^2 + 49a + 4a + 28 = 0$

Step6: Factor by grouping

$7a(a+7) + 4(a+7) = 0$

Step7: Extract common binomial factor

$(7a+4)(a+7) = 0$

Step8: Solve for $a$ using zero product rule

Set each factor to 0:
$7a+4=0 \implies a=-\frac{4}{7}$; $a+7=0 \implies a=-7$

Answer:

Factored form (Problem 3): $(2v+1)(v+5)$
Solutions (Problem 4): $a = -7$, $a = -\frac{4}{7}$
Factors (Problem 4): $(7a+4)(a+7)$