Holly Guerrero

2021-12-28

${u}^{2}+7u+6$

Paul Mitchell

Step 1
Given: ${u}^{2}+7u+6$
We have to find integers b and d such that
${u}^{2}+7u+6=\left(u+b\right)\left(u+d\right)$
$={u}^{2}+du+bu+bd$
${u}^{2}+7u+6={u}^{2}+\left(b+d\right)u+bd$
SInce: $bd=6$ b and d are factors of 6
Similarly: $b+d=7$
Table:

To check:
$\left(u+1\right)\left(u+6\right)=u\left(u+6\right)+1\left(u+6\right)$
${u}^{2}+6u+u+6$
$={u}^{2}+7y+6$

einfachmoipf

Step 1
Let's factor ${u}^{2}+7u+6$
The middle number is 7 and the last number is 6.
Factoring means we want something like
$\left(u+\underset{―}{}\right)\left(u+\underset{―}{}\right)$
Which numbers go in the blanks?
We need two numbers that
Multiply together to get 6
Can you think of the two numbers?
Try 1 and 6
$1+6=7$
$1×6=6$
Fill in the blanks in
$\left(u+\underset{―}{}\right)\left(u+\underset{―}{}\right)$
with 1 and 6 to get
$\left(u+1\right)\left(u+6\right)$

user_27qwe

Step 1
Quadratic polynomial can be factored using the transformation $a{x}^{2}+bx+c=a\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)$, where ${x}_{1}$ and ${x}_{2}$ are the solutions of the quadratic equation $a{x}^{2}+bx+c=0$
${u}^{2}+7u+6=0$
All equations of the form $a{x}^{2}+bx+c=0$ can be solved using the quadratic formula $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$u=\frac{-7±\sqrt{{7}^{2}-4×6}}{2}$
Square 7
$u=\frac{-7±\sqrt{49-4×6}}{2}$
Multiply -4 times 6.
$u=\frac{-7±\sqrt{49-24}}{2}$
$u=\frac{-7±\sqrt{25}}{2}$
Take the square root of 25.
$u=\frac{-7±5}{2}$
Now solve the equation $u=\frac{-7±5}{2}$ when $±$ is plus. Add -7 to 5
$u=\frac{-2}{2}$
Divide -2 by 2
u=-1
Now solve the equation $u=\frac{-7±5}{2}$ when $±$ is minus. Subtract 5 from -7
u=$\frac{-12}{2}$
Divide -12 by 2
u=-6
Factor the original expression using ax${}^{2}$+bx+c=a(x-x${}_{1}$)(x-x${}_{2}$). Substitute -1 for x${}_{1}$ and -6 for x${}_{2}$
u${}^{2}$+7u+6=(u-(-1))(u-(-6))
Simplify all the expressions of the form p-(-q) to p+q
u${}^{2}$+7u+6=(u+1)(u+6)
(u+1)(u+6)

Do you have a similar question?