Solve (-2p^ 2 +10p-14)+(-9p^ 2 +4p)|1 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$(-2p^{2}+10p-14)+(-9p^{2}+4p)|1$$

Evaluate

$-11p^{2}+14p-14$

Solution Steps

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $1$ is $1$.

$$-2p^{2}+10p-14+\left(-9p^{2}+4p\right)\times 1$$

Use the distributive property to multiply $-9p^{2}+4p$ by $1$.

$$-2p^{2}+10p-14-9p^{2}+4p$$

Combine $-2p^{2}$ and $-9p^{2}$ to get $-11p^{2}$.

$$-11p^{2}+10p-14+4p$$

Combine $10p$ and $4p$ to get $14p$.

$$-11p^{2}+14p-14$$

Differentiate w.r.t. p

$14-22p$

Steps Using Derivative Rule for Sum

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $1$ is $1$.

$$\frac{\mathrm{d}}{\mathrm{d}p}(-2p^{2}+10p-14+\left(-9p^{2}+4p\right)\times 1)$$

Use the distributive property to multiply $-9p^{2}+4p$ by $1$.

$$\frac{\mathrm{d}}{\mathrm{d}p}(-2p^{2}+10p-14-9p^{2}+4p)$$

Combine $-2p^{2}$ and $-9p^{2}$ to get $-11p^{2}$.

$$\frac{\mathrm{d}}{\mathrm{d}p}(-11p^{2}+10p-14+4p)$$

Combine $10p$ and $4p$ to get $14p$.

$$\frac{\mathrm{d}}{\mathrm{d}p}(-11p^{2}+14p-14)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is $0$. The derivative of $ax^{n}$ is $nax^{n-1}$.

$$2\left(-11\right)p^{2-1}+14p^{1-1}$$

Multiply $2$ times $-11$.

$$-22p^{2-1}+14p^{1-1}$$

Subtract $1$ from $2$.

$$-22p^{1}+14p^{1-1}$$

Subtract $1$ from $1$.

$$-22p^{1}+14p^{0}$$

For any term $t$, $t^{1}=t$.

$$-22p+14p^{0}$$

For any term $t$ except $0$, $t^{0}=1$.

$$-22p+14\times 1$$

For any term $t$, $t\times 1=t$ and $1t=t$.

$$-22p+14$$