Use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ to find the following product.
$(x + 3) (x + 7)$

x^{2} - 10 x + 21

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + 10 x + 21

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} - 10 x - 21

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + 2 x + 21

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} + 10 x + 21$
Given: $(x + 3) (x + 7)$

$= x^{2} + (3 + 7) x + 3 \times 7$

$= x^{2} + 10 x + 21$

Suggest Corrections

Similar questions

Q. Solve the equation

∣ ∣ ∣ \frac{x^{2} - 8 x + 12}{x^{2} - 10 x + 21} ∣ ∣ ∣ = - \frac{x^{2} - 8 x + 12}{x^{2} - 10 x + 21}

, the ans is

x \in [2, a) \cup [b, 7)

Find b-a?

Q. The number of integral solutions of the equations

∣ ∣ ∣ \frac{x^{2} - 10 x - 21}{x^{2} - 12 x + 32} ∣ ∣ ∣ = - (\frac{x^{2} - 10 x + 21}{x^{2} - 12 x + 32})

is(are)

Q. Find the following products using the identity : (x+a)(x+b)=

x^{2} + (a + b) x + a b .

1. (x+2)(x+3)
2.(x-10)(x+8)
3.(x+15)(x-6)
4. (x+7)(x+7)

Q. Factorize the following trinomial:

x^{2} + 10 x + 21

Let f(x) = $x^{2}$ -1, 0<x<2
and 2x+3, 2 $\leq$ x<3, The quadratic equation whose roots are ${lim}_{x \to 2^{-}} f (x), a n d {lim}_{x \to 2^{+}} f (x)$

Join BYJU'S Learning Program

Use the identity (x+a)(x+b)=x2+(a+b)x+ab to find the following product.(x+3)(x+7)

The correct option is B x2+10x+21Given: (x+3)(x+7)=x2+(3+7)x+3×7=x2+10x+21

Use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ to find the following product.
$(x + 3) (x + 7)$

The correct option is B $x^{2} + 10 x + 21$
Given: $(x + 3) (x + 7)$

$= x^{2} + (3 + 7) x + 3 \times 7$

$= x^{2} + 10 x + 21$